Mixing
$f+f'+f''geq0$,Prove the $f$ has a lower bound
$begingroup$
Let $fin C^2(a,b)$ such that $f+f'+f''geq0$
Prove that $f$ has a lower bound.
$Myquad Attempt$
$1.quad$Suppose that $f$ has no lower bound at x=b,so there is a sequence${x_n}$,which converges to b($lim_{ntoinfty}{x_n}=b$),and $f'(x_n)<0,f(x_n)<-n$
$2.quadforall kin mathbb N,exists Ninmathbb N,n>N,x_n>x_k$,then prove that $$int_{{x|xin(x_k,x_n)land f(x)>0}}{}f^2(x)dxleq C$$
$3.quad$Prove: $$0leqint_{x_k}^{x_n}{(f+f'+f'')}dxleq0+f(x_k)-f(x_n)+Cto-infty$$
$quadquadquadquadquadquadquad$This contradicts the problem
So that's my idea，but I can't do it from step 2.And my idea might be wrong.
Edit in 2019/2/16
I solve the question that if $q=0$.There is some Chinese characters in my answer.I hope it doesn't bother you
real-analysis contest-math
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
$endgroup$
|
show 2 more comments
$begingroup$
Let $fin C^2(a,b)$ such that $f+f'+f''geq0$
Prove that $f$ has a lower bound.
$Myquad Attempt$
$1.quad$Suppose that $f$ has no lower bound at x=b,so there is a sequence${x_n}$,which converges to b($lim_{ntoinfty}{x_n}=b$),and $f'(x_n)<0,f(x_n)<-n$
$2.quadforall kin mathbb N,exists Ninmathbb N,n>N,x_n>x_k$,then prove that $$int_{{x|xin(x_k,x_n)land f(x)>0}}{}f^2(x)dxleq C$$
$3.quad$Prove: $$0leqint_{x_k}^{x_n}{(f+f'+f'')}dxleq0+f(x_k)-f(x_n)+Cto-infty$$
$quadquadquadquadquadquadquad$This contradicts the problem
So that's my idea，but I can't do it from step 2.And my idea might be wrong.
Edit in 2019/2/16
I solve the question that if $q=0$.There is some Chinese characters in my answer.I hope it doesn't bother you
real-analysis contest-math
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
$endgroup$
1
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
1
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46
|
show 2 more comments
$begingroup$
Let $fin C^2(a,b)$ such that $f+f'+f''geq0$
Prove that $f$ has a lower bound.
$Myquad Attempt$
$1.quad$Suppose that $f$ has no lower bound at x=b,so there is a sequence${x_n}$,which converges to b($lim_{ntoinfty}{x_n}=b$),and $f'(x_n)<0,f(x_n)<-n$
$2.quadforall kin mathbb N,exists Ninmathbb N,n>N,x_n>x_k$,then prove that $$int_{{x|xin(x_k,x_n)land f(x)>0}}{}f^2(x)dxleq C$$
$3.quad$Prove: $$0leqint_{x_k}^{x_n}{(f+f'+f'')}dxleq0+f(x_k)-f(x_n)+Cto-infty$$
$quadquadquadquadquadquadquad$This contradicts the problem
So that's my idea，but I can't do it from step 2.And my idea might be wrong.
Edit in 2019/2/16
I solve the question that if $q=0$.There is some Chinese characters in my answer.I hope it doesn't bother you
real-analysis contest-math
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
$endgroup$
Let $fin C^2(a,b)$ such that $f+f'+f''geq0$
Prove that $f$ has a lower bound.
$Myquad Attempt$
$1.quad$Suppose that $f$ has no lower bound at x=b,so there is a sequence${x_n}$,which converges to b($lim_{ntoinfty}{x_n}=b$),and $f'(x_n)<0,f(x_n)<-n$
$2.quadforall kin mathbb N,exists Ninmathbb N,n>N,x_n>x_k$,then prove that $$int_{{x|xin(x_k,x_n)land f(x)>0}}{}f^2(x)dxleq C$$
$3.quad$Prove: $$0leqint_{x_k}^{x_n}{(f+f'+f'')}dxleq0+f(x_k)-f(x_n)+Cto-infty$$
$quadquadquadquadquadquadquad$This contradicts the problem
So that's my idea，but I can't do it from step 2.And my idea might be wrong.
Edit in 2019/2/16
I solve the question that if $q=0$.There is some Chinese characters in my answer.I hope it doesn't bother you
real-analysis contest-math
real-analysis contest-math
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
edited Feb 16 at 2:31
LiTaichi
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
asked Jan 20 at 3:05
LiTaichiLiTaichi
3587
3587
1
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
1
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46
|
show 2 more comments
1
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
1
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46
1
1
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
1
1
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Lemma. Given $a<b,$ any $gin C^2(a,b)$ satisfying $g''+ggeq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $max(a,b-pi)<t<s<b$ define $h_s(t)=g(t)cos(t-s)-g'(t)sin(t-s).$ Then
$$h'_s(t)=(g(t)cos(t-s)-g'(t)sin(t-s))'=-(g(t)+g''(t))sin(t-s)geq 0$$
because $sin(t-s)<0.$ So
$$g(s)=h_s(s)geq h_s(t)geq -|g(t)|-|g'(t)|.$$
Picking any $max(a,b-pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $Gin C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $fin C^2(a,b)$ satisfying $f''+f'+fgeq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/sqrt 3}f(tfrac{2}{sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/sqrt 3}(tfrac{4}{3}f''(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f'(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f(tfrac{2}{sqrt 3}t))geq 0$$
for $a<tfrac{2}{sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+ggeq 0$ on $(tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)geq C$ for all $tin (tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(tfrac{sqrt3}2t)geq e^{a/2}C.$$
answered Feb 15 at 20:38
DapDap
17.7k841
$endgroup$
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
|
show 1 more comment
$begingroup$
Proof: Let $F(x) = e^{frac{x}{2}}f(x)$,then
$$begin{align}frac{3}{4}F(x)+F''(x) &=frac{3}{4} e^{frac{x}{2}}f(x)+e^{frac{x}{2}}(frac{1}{4}f(x)+f'(x)+f''(x))\ &=e^{frac{x}{2}}(f(x)+f'(x)+f''(x))\ &geq0 end{align}$$
(1) $F(x)<0,forall xin(a,b).$
$F''(x)geq -F(x)geq 0$, so $F(x)$ is convex function. Picking any $x_0in(a,b)$,we have
$$F(x)geq F(x_0)+F'(x_0)(x-x_0),quadquadquadquadforall xin(a,b).$$
(2) $exists x_0in(a,b),F(x_0)geq0$
We suppose that there's $n$ zero points in $(a,b)$ at most, and $2leq n<infty$.
For $0leq nleq 1$ , we assume that $F(x)<0 ,xin(a,x_0),F(x)geq 0,xin[x_0,b),$
In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1
In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $ngeq2$, $exists alpha,betain(a,b),F(alpha)=F(beta)=0$, and
$$F(x)geq 0, quad quadquadquadquadquad xin(alpha,beta).$$
Let $gammain(alpha,beta)$ is absolute maximum point of $F(x)$ in $[alpha,beta]$, then for $t>gamma,F''(t)leq0$.
Then $exists sin[gamma,t],F'(s)=0$,
$$F'(x)<0,quad quadquadquadquadquad xin(s,t)$$
So $|F'(x)|^2+frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
$$begin{align}F'(x)&geq-sqrt{|F'(s)|^2+frac{3}{4}|F(s)|^2-frac{3}{4}|F(x)|^2}\ &=-frac{3}{4}sqrt{|F(s)|^2-|F(x)|^2}\ &geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin(s,t) end{align}$$
We get,
$$F'(x)geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin[gamma,beta]$$
that is, $arcsinfrac{F'(x)}{F'(gamma)}+frac{3}{4}x $ is monotone increasing in $[gamma,beta]$, then
$$
frac{3beta}{4}=arcsinfrac{F'(beta)}{F'(gamma)}+frac{3}{4}xgeq arcsinfrac{F'(gamma)}{F'(gamma)}+frac{3}{4}x=frac{pi}{2}+frac{3gamma}{4}
$$
which implies $beta-alphageqbeta-gammageqfrac{2pi}{3}$, we get $2leq n<infty$
If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,xin[x_1,b)$,
$$F(x)geq F'(x_1)(x-x_1),quad quadforall xin[x_1,b).$$
We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
Obviously, $f(x)$ has a lower pound in $(a,b)$.
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080121%2ffff-geq0-prove-the-f-has-a-lower-bound%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Lemma. Given $a<b,$ any $gin C^2(a,b)$ satisfying $g''+ggeq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $max(a,b-pi)<t<s<b$ define $h_s(t)=g(t)cos(t-s)-g'(t)sin(t-s).$ Then
$$h'_s(t)=(g(t)cos(t-s)-g'(t)sin(t-s))'=-(g(t)+g''(t))sin(t-s)geq 0$$
because $sin(t-s)<0.$ So
$$g(s)=h_s(s)geq h_s(t)geq -|g(t)|-|g'(t)|.$$
Picking any $max(a,b-pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $Gin C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $fin C^2(a,b)$ satisfying $f''+f'+fgeq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/sqrt 3}f(tfrac{2}{sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/sqrt 3}(tfrac{4}{3}f''(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f'(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f(tfrac{2}{sqrt 3}t))geq 0$$
for $a<tfrac{2}{sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+ggeq 0$ on $(tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)geq C$ for all $tin (tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(tfrac{sqrt3}2t)geq e^{a/2}C.$$
answered Feb 15 at 20:38
DapDap
17.7k841
$endgroup$
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
|
show 1 more comment
$begingroup$
Lemma. Given $a<b,$ any $gin C^2(a,b)$ satisfying $g''+ggeq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $max(a,b-pi)<t<s<b$ define $h_s(t)=g(t)cos(t-s)-g'(t)sin(t-s).$ Then
$$h'_s(t)=(g(t)cos(t-s)-g'(t)sin(t-s))'=-(g(t)+g''(t))sin(t-s)geq 0$$
because $sin(t-s)<0.$ So
$$g(s)=h_s(s)geq h_s(t)geq -|g(t)|-|g'(t)|.$$
Picking any $max(a,b-pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $Gin C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $fin C^2(a,b)$ satisfying $f''+f'+fgeq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/sqrt 3}f(tfrac{2}{sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/sqrt 3}(tfrac{4}{3}f''(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f'(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f(tfrac{2}{sqrt 3}t))geq 0$$
for $a<tfrac{2}{sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+ggeq 0$ on $(tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)geq C$ for all $tin (tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(tfrac{sqrt3}2t)geq e^{a/2}C.$$
answered Feb 15 at 20:38
DapDap
17.7k841
$endgroup$
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
|
show 1 more comment
$begingroup$
Lemma. Given $a<b,$ any $gin C^2(a,b)$ satisfying $g''+ggeq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $max(a,b-pi)<t<s<b$ define $h_s(t)=g(t)cos(t-s)-g'(t)sin(t-s).$ Then
$$h'_s(t)=(g(t)cos(t-s)-g'(t)sin(t-s))'=-(g(t)+g''(t))sin(t-s)geq 0$$
because $sin(t-s)<0.$ So
$$g(s)=h_s(s)geq h_s(t)geq -|g(t)|-|g'(t)|.$$
Picking any $max(a,b-pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $Gin C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $fin C^2(a,b)$ satisfying $f''+f'+fgeq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/sqrt 3}f(tfrac{2}{sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/sqrt 3}(tfrac{4}{3}f''(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f'(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f(tfrac{2}{sqrt 3}t))geq 0$$
for $a<tfrac{2}{sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+ggeq 0$ on $(tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)geq C$ for all $tin (tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(tfrac{sqrt3}2t)geq e^{a/2}C.$$
answered Feb 15 at 20:38
DapDap
17.7k841
$endgroup$
Lemma. Given $a<b,$ any $gin C^2(a,b)$ satisfying $g''+ggeq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $max(a,b-pi)<t<s<b$ define $h_s(t)=g(t)cos(t-s)-g'(t)sin(t-s).$ Then
$$h'_s(t)=(g(t)cos(t-s)-g'(t)sin(t-s))'=-(g(t)+g''(t))sin(t-s)geq 0$$
because $sin(t-s)<0.$ So
$$g(s)=h_s(s)geq h_s(t)geq -|g(t)|-|g'(t)|.$$
Picking any $max(a,b-pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $Gin C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $fin C^2(a,b)$ satisfying $f''+f'+fgeq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/sqrt 3}f(tfrac{2}{sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/sqrt 3}(tfrac{4}{3}f''(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f'(tfrac{2}{sqrt 3}t)+tfrac{4}{3}f(tfrac{2}{sqrt 3}t))geq 0$$
for $a<tfrac{2}{sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+ggeq 0$ on $(tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)geq C$ for all $tin (tfrac{sqrt 3}{2}a,tfrac{sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(tfrac{sqrt3}2t)geq e^{a/2}C.$$
answered Feb 15 at 20:38
DapDap
17.7k841
edited Feb 18 at 6:42
answered Feb 15 at 20:38
DapDap
17.7k841
answered Feb 15 at 20:38
DapDap
17.7k841
answered Feb 15 at 20:38
DapDap
17.7k841
17.7k841
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
|
show 1 more comment
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@Dap Your auxiliary function $h_{s}$ is like a stroke of genius to me. :) Could you give me some references about how to apply this Picone identity?
$endgroup$
– Eric Yau
Feb 16 at 7:48
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@EricYau: this is a variant of the Sturm comparison theorem. The most general statement I know is "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 6.1.
$endgroup$
– Dap
Feb 16 at 10:13
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap But for $(a,t)$,does $f$ has a lower bound?
$endgroup$
– LiTaichi
Feb 16 at 12:06
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@Dap the $g(x)$ is bounded and $f(x)$ is bounded are equivalent?But I still don't think your proof can prove $f$ has a lower bound in$ (a,b)? Could you please write more detail?
$endgroup$
– LiTaichi
Feb 17 at 2:31
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
$begingroup$
@梦里年华似烟花: ok, I've added some detail and made the structure of the argument explicit
$endgroup$
– Dap
Feb 18 at 6:43
|
show 1 more comment
$begingroup$
Proof: Let $F(x) = e^{frac{x}{2}}f(x)$,then
$$begin{align}frac{3}{4}F(x)+F''(x) &=frac{3}{4} e^{frac{x}{2}}f(x)+e^{frac{x}{2}}(frac{1}{4}f(x)+f'(x)+f''(x))\ &=e^{frac{x}{2}}(f(x)+f'(x)+f''(x))\ &geq0 end{align}$$
(1) $F(x)<0,forall xin(a,b).$
$F''(x)geq -F(x)geq 0$, so $F(x)$ is convex function. Picking any $x_0in(a,b)$,we have
$$F(x)geq F(x_0)+F'(x_0)(x-x_0),quadquadquadquadforall xin(a,b).$$
(2) $exists x_0in(a,b),F(x_0)geq0$
We suppose that there's $n$ zero points in $(a,b)$ at most, and $2leq n<infty$.
For $0leq nleq 1$ , we assume that $F(x)<0 ,xin(a,x_0),F(x)geq 0,xin[x_0,b),$
In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1
In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $ngeq2$, $exists alpha,betain(a,b),F(alpha)=F(beta)=0$, and
$$F(x)geq 0, quad quadquadquadquadquad xin(alpha,beta).$$
Let $gammain(alpha,beta)$ is absolute maximum point of $F(x)$ in $[alpha,beta]$, then for $t>gamma,F''(t)leq0$.
Then $exists sin[gamma,t],F'(s)=0$,
$$F'(x)<0,quad quadquadquadquadquad xin(s,t)$$
So $|F'(x)|^2+frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
$$begin{align}F'(x)&geq-sqrt{|F'(s)|^2+frac{3}{4}|F(s)|^2-frac{3}{4}|F(x)|^2}\ &=-frac{3}{4}sqrt{|F(s)|^2-|F(x)|^2}\ &geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin(s,t) end{align}$$
We get,
$$F'(x)geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin[gamma,beta]$$
that is, $arcsinfrac{F'(x)}{F'(gamma)}+frac{3}{4}x $ is monotone increasing in $[gamma,beta]$, then
$$
frac{3beta}{4}=arcsinfrac{F'(beta)}{F'(gamma)}+frac{3}{4}xgeq arcsinfrac{F'(gamma)}{F'(gamma)}+frac{3}{4}x=frac{pi}{2}+frac{3gamma}{4}
$$
which implies $beta-alphageqbeta-gammageqfrac{2pi}{3}$, we get $2leq n<infty$
If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,xin[x_1,b)$,
$$F(x)geq F'(x_1)(x-x_1),quad quadforall xin[x_1,b).$$
We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
Obviously, $f(x)$ has a lower pound in $(a,b)$.
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
$endgroup$
add a comment |
$begingroup$
Proof: Let $F(x) = e^{frac{x}{2}}f(x)$,then
$$begin{align}frac{3}{4}F(x)+F''(x) &=frac{3}{4} e^{frac{x}{2}}f(x)+e^{frac{x}{2}}(frac{1}{4}f(x)+f'(x)+f''(x))\ &=e^{frac{x}{2}}(f(x)+f'(x)+f''(x))\ &geq0 end{align}$$
(1) $F(x)<0,forall xin(a,b).$
$F''(x)geq -F(x)geq 0$, so $F(x)$ is convex function. Picking any $x_0in(a,b)$,we have
$$F(x)geq F(x_0)+F'(x_0)(x-x_0),quadquadquadquadforall xin(a,b).$$
(2) $exists x_0in(a,b),F(x_0)geq0$
We suppose that there's $n$ zero points in $(a,b)$ at most, and $2leq n<infty$.
For $0leq nleq 1$ , we assume that $F(x)<0 ,xin(a,x_0),F(x)geq 0,xin[x_0,b),$
In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1
In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $ngeq2$, $exists alpha,betain(a,b),F(alpha)=F(beta)=0$, and
$$F(x)geq 0, quad quadquadquadquadquad xin(alpha,beta).$$
Let $gammain(alpha,beta)$ is absolute maximum point of $F(x)$ in $[alpha,beta]$, then for $t>gamma,F''(t)leq0$.
Then $exists sin[gamma,t],F'(s)=0$,
$$F'(x)<0,quad quadquadquadquadquad xin(s,t)$$
So $|F'(x)|^2+frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
$$begin{align}F'(x)&geq-sqrt{|F'(s)|^2+frac{3}{4}|F(s)|^2-frac{3}{4}|F(x)|^2}\ &=-frac{3}{4}sqrt{|F(s)|^2-|F(x)|^2}\ &geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin(s,t) end{align}$$
We get,
$$F'(x)geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin[gamma,beta]$$
that is, $arcsinfrac{F'(x)}{F'(gamma)}+frac{3}{4}x $ is monotone increasing in $[gamma,beta]$, then
$$
frac{3beta}{4}=arcsinfrac{F'(beta)}{F'(gamma)}+frac{3}{4}xgeq arcsinfrac{F'(gamma)}{F'(gamma)}+frac{3}{4}x=frac{pi}{2}+frac{3gamma}{4}
$$
which implies $beta-alphageqbeta-gammageqfrac{2pi}{3}$, we get $2leq n<infty$
If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,xin[x_1,b)$,
$$F(x)geq F'(x_1)(x-x_1),quad quadforall xin[x_1,b).$$
We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
Obviously, $f(x)$ has a lower pound in $(a,b)$.
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
$endgroup$
add a comment |
$begingroup$
Proof: Let $F(x) = e^{frac{x}{2}}f(x)$,then
$$begin{align}frac{3}{4}F(x)+F''(x) &=frac{3}{4} e^{frac{x}{2}}f(x)+e^{frac{x}{2}}(frac{1}{4}f(x)+f'(x)+f''(x))\ &=e^{frac{x}{2}}(f(x)+f'(x)+f''(x))\ &geq0 end{align}$$
(1) $F(x)<0,forall xin(a,b).$
$F''(x)geq -F(x)geq 0$, so $F(x)$ is convex function. Picking any $x_0in(a,b)$,we have
$$F(x)geq F(x_0)+F'(x_0)(x-x_0),quadquadquadquadforall xin(a,b).$$
(2) $exists x_0in(a,b),F(x_0)geq0$
We suppose that there's $n$ zero points in $(a,b)$ at most, and $2leq n<infty$.
For $0leq nleq 1$ , we assume that $F(x)<0 ,xin(a,x_0),F(x)geq 0,xin[x_0,b),$
In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1
In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $ngeq2$, $exists alpha,betain(a,b),F(alpha)=F(beta)=0$, and
$$F(x)geq 0, quad quadquadquadquadquad xin(alpha,beta).$$
Let $gammain(alpha,beta)$ is absolute maximum point of $F(x)$ in $[alpha,beta]$, then for $t>gamma,F''(t)leq0$.
Then $exists sin[gamma,t],F'(s)=0$,
$$F'(x)<0,quad quadquadquadquadquad xin(s,t)$$
So $|F'(x)|^2+frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
$$begin{align}F'(x)&geq-sqrt{|F'(s)|^2+frac{3}{4}|F(s)|^2-frac{3}{4}|F(x)|^2}\ &=-frac{3}{4}sqrt{|F(s)|^2-|F(x)|^2}\ &geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin(s,t) end{align}$$
We get,
$$F'(x)geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin[gamma,beta]$$
that is, $arcsinfrac{F'(x)}{F'(gamma)}+frac{3}{4}x $ is monotone increasing in $[gamma,beta]$, then
$$
frac{3beta}{4}=arcsinfrac{F'(beta)}{F'(gamma)}+frac{3}{4}xgeq arcsinfrac{F'(gamma)}{F'(gamma)}+frac{3}{4}x=frac{pi}{2}+frac{3gamma}{4}
$$
which implies $beta-alphageqbeta-gammageqfrac{2pi}{3}$, we get $2leq n<infty$
If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,xin[x_1,b)$,
$$F(x)geq F'(x_1)(x-x_1),quad quadforall xin[x_1,b).$$
We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
Obviously, $f(x)$ has a lower pound in $(a,b)$.
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
$endgroup$
Proof: Let $F(x) = e^{frac{x}{2}}f(x)$,then
$$begin{align}frac{3}{4}F(x)+F''(x) &=frac{3}{4} e^{frac{x}{2}}f(x)+e^{frac{x}{2}}(frac{1}{4}f(x)+f'(x)+f''(x))\ &=e^{frac{x}{2}}(f(x)+f'(x)+f''(x))\ &geq0 end{align}$$
(1) $F(x)<0,forall xin(a,b).$
$F''(x)geq -F(x)geq 0$, so $F(x)$ is convex function. Picking any $x_0in(a,b)$,we have
$$F(x)geq F(x_0)+F'(x_0)(x-x_0),quadquadquadquadforall xin(a,b).$$
(2) $exists x_0in(a,b),F(x_0)geq0$
We suppose that there's $n$ zero points in $(a,b)$ at most, and $2leq n<infty$.
For $0leq nleq 1$ , we assume that $F(x)<0 ,xin(a,x_0),F(x)geq 0,xin[x_0,b),$
In $(a,x_0)$, we know that $F(x)$ has a lower bound from 1
In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $ngeq2$, $exists alpha,betain(a,b),F(alpha)=F(beta)=0$, and
$$F(x)geq 0, quad quadquadquadquadquad xin(alpha,beta).$$
Let $gammain(alpha,beta)$ is absolute maximum point of $F(x)$ in $[alpha,beta]$, then for $t>gamma,F''(t)leq0$.
Then $exists sin[gamma,t],F'(s)=0$,
$$F'(x)<0,quad quadquadquadquadquad xin(s,t)$$
So $|F'(x)|^2+frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
$$begin{align}F'(x)&geq-sqrt{|F'(s)|^2+frac{3}{4}|F(s)|^2-frac{3}{4}|F(x)|^2}\ &=-frac{3}{4}sqrt{|F(s)|^2-|F(x)|^2}\ &geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin(s,t) end{align}$$
We get,
$$F'(x)geq-frac{3}{4}sqrt{|F(gamma)|^2-|F(x)|^2},quad xin[gamma,beta]$$
that is, $arcsinfrac{F'(x)}{F'(gamma)}+frac{3}{4}x $ is monotone increasing in $[gamma,beta]$, then
$$
frac{3beta}{4}=arcsinfrac{F'(beta)}{F'(gamma)}+frac{3}{4}xgeq arcsinfrac{F'(gamma)}{F'(gamma)}+frac{3}{4}x=frac{pi}{2}+frac{3gamma}{4}
$$
which implies $beta-alphageqbeta-gammageqfrac{2pi}{3}$, we get $2leq n<infty$
If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,xin[x_1,b)$,
$$F(x)geq F'(x_1)(x-x_1),quad quadforall xin[x_1,b).$$
We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
Obviously, $f(x)$ has a lower pound in $(a,b)$.
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
answered Feb 20 at 2:09
LiTaichiLiTaichi
3587
3587
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080121%2ffff-geq0-prove-the-f-has-a-lower-bound%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
I'm the question writer.I'm sorry,the question should prove the function has a lower bound
$endgroup$
– LiTaichi
Jan 20 at 3:09
$begingroup$
Please then edit your question to say this; otherwise, you confuse folks (like Yours Truly for example). Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 3:29
$begingroup$
Please put what you have tried in the question body with an edit.
$endgroup$
– YiFan
Jan 20 at 3:33
1
$begingroup$
For these type of questions, it seems like multiplying by $e^{ax}$ for some magic value of $a$ seems to work.
$endgroup$
– marty cohen
Jan 20 at 6:25
$begingroup$
$f+f'+f''leq0$,if$f'leq0$,$f+f'+f''leq-(f')^2leq0$,so $[f^2+f'^2]'leq0$,this might be a useful function,I saw a similar problem using this function
$endgroup$
– LiTaichi
Jan 20 at 7:46