Mixing
If the right derivative is positive at each point, can I conclude that $f$ is increasing at right of each...
$begingroup$
Let $f:mathbb Rto mathbb R$ a derivable function at right of each point, i.e. $$lim_{hto 0^+}frac{f(x+h)-f(x)}{h}=f_d'(x)$$
exist for all $x$. We suppose that $f_d'(x)geq 0$ for all $x$. Can we conclude that $f$ is increasing at right of each point ? i.e for all $x$, there is $delta =delta_x $ s.t. $f(x)leq f(y)$ for all $yin [x, x+delta ]$ ?
I tried as follow :
$$f(x+h)=f(x)+f'_d(x)h+hvarepsilon (h)$$
for $h>0$ where $varepsilon (h)to 0$ when $hto 0^+$. In particular, $$f(x+h)geq f(x)+hvarepsilon (h).$$
Now, I don't see why this would prove the statement since $varepsilon (h)$ could be negative. Any idea ?
real-analysis
asked Jan 26 at 17:40
DylanDylan
3118
$endgroup$
|
show 4 more comments
$begingroup$
Let $f:mathbb Rto mathbb R$ a derivable function at right of each point, i.e. $$lim_{hto 0^+}frac{f(x+h)-f(x)}{h}=f_d'(x)$$
exist for all $x$. We suppose that $f_d'(x)geq 0$ for all $x$. Can we conclude that $f$ is increasing at right of each point ? i.e for all $x$, there is $delta =delta_x $ s.t. $f(x)leq f(y)$ for all $yin [x, x+delta ]$ ?
I tried as follow :
$$f(x+h)=f(x)+f'_d(x)h+hvarepsilon (h)$$
for $h>0$ where $varepsilon (h)to 0$ when $hto 0^+$. In particular, $$f(x+h)geq f(x)+hvarepsilon (h).$$
Now, I don't see why this would prove the statement since $varepsilon (h)$ could be negative. Any idea ?
real-analysis
asked Jan 26 at 17:40
DylanDylan
3118
$endgroup$
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24
|
show 4 more comments
$begingroup$
Let $f:mathbb Rto mathbb R$ a derivable function at right of each point, i.e. $$lim_{hto 0^+}frac{f(x+h)-f(x)}{h}=f_d'(x)$$
exist for all $x$. We suppose that $f_d'(x)geq 0$ for all $x$. Can we conclude that $f$ is increasing at right of each point ? i.e for all $x$, there is $delta =delta_x $ s.t. $f(x)leq f(y)$ for all $yin [x, x+delta ]$ ?
I tried as follow :
$$f(x+h)=f(x)+f'_d(x)h+hvarepsilon (h)$$
for $h>0$ where $varepsilon (h)to 0$ when $hto 0^+$. In particular, $$f(x+h)geq f(x)+hvarepsilon (h).$$
Now, I don't see why this would prove the statement since $varepsilon (h)$ could be negative. Any idea ?
real-analysis
asked Jan 26 at 17:40
DylanDylan
3118
$endgroup$
Let $f:mathbb Rto mathbb R$ a derivable function at right of each point, i.e. $$lim_{hto 0^+}frac{f(x+h)-f(x)}{h}=f_d'(x)$$
exist for all $x$. We suppose that $f_d'(x)geq 0$ for all $x$. Can we conclude that $f$ is increasing at right of each point ? i.e for all $x$, there is $delta =delta_x $ s.t. $f(x)leq f(y)$ for all $yin [x, x+delta ]$ ?
I tried as follow :
$$f(x+h)=f(x)+f'_d(x)h+hvarepsilon (h)$$
for $h>0$ where $varepsilon (h)to 0$ when $hto 0^+$. In particular, $$f(x+h)geq f(x)+hvarepsilon (h).$$
Now, I don't see why this would prove the statement since $varepsilon (h)$ could be negative. Any idea ?
real-analysis
real-analysis
asked Jan 26 at 17:40
DylanDylan
3118
asked Jan 26 at 17:40
DylanDylan
3118
edited Jan 26 at 17:56
Dylan
asked Jan 26 at 17:40
DylanDylan
3118
asked Jan 26 at 17:40
DylanDylan
3118
asked Jan 26 at 17:40
DylanDylan
3118
3118
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24
|
show 4 more comments
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here is a counterexample. Define
$$
f(x) = begin{cases}
0 & (x leq 0)\
-frac{1}{2^n} & left(frac{1}{sqrt{2^n}} leq x < frac{1}{sqrt{2^{n-1}}}; n > 1 right)\
-frac{1}{2} & left(xgeq frac{1}{sqrt 2}right)
end{cases}
$$
Clearly
$$f'_+(x) = lim_{hrightarrow 0^+}frac{f(x+h)-f(x)}{h}=0$$
for all $x neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 leq f(x) leq 0,$$
so that
$$-x leq frac{f(x)}{x} leq 0.$$
Thus
$$f'_+(0) = lim_{hrightarrow 0^+} frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) geq 0$, for all $xin Bbb R$, but for any $delta > 0$, $xin (0,delta) Rightarrow f(x) < 0$.
In conclusion, if $f(x)$ is not continuous your statement is clearly false.
EDIT
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = begin{cases}frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x+sqrt[4]{2^{-k+1}})- sqrt{2^{-k+1}} & left(-sqrt[4]{2^{-k+1}}leq x<-sqrt[4]{2^{-k}}; kin Bbb Zright)\ 0 & (x=0)\ frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x-sqrt[4]{2^{-k+1}})+ sqrt{2^{-k+1}}& left(sqrt[4]{2^{-k}}leq x<sqrt[4]{2^{-k+1}}; kin Bbb Zright) .end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.
answered Feb 2 at 17:09
MatteoMatteo
1,142313
$endgroup$
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
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%2f3088518%2fif-the-right-derivative-is-positive-at-each-point-can-i-conclude-that-f-is-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a counterexample. Define
$$
f(x) = begin{cases}
0 & (x leq 0)\
-frac{1}{2^n} & left(frac{1}{sqrt{2^n}} leq x < frac{1}{sqrt{2^{n-1}}}; n > 1 right)\
-frac{1}{2} & left(xgeq frac{1}{sqrt 2}right)
end{cases}
$$
Clearly
$$f'_+(x) = lim_{hrightarrow 0^+}frac{f(x+h)-f(x)}{h}=0$$
for all $x neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 leq f(x) leq 0,$$
so that
$$-x leq frac{f(x)}{x} leq 0.$$
Thus
$$f'_+(0) = lim_{hrightarrow 0^+} frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) geq 0$, for all $xin Bbb R$, but for any $delta > 0$, $xin (0,delta) Rightarrow f(x) < 0$.
In conclusion, if $f(x)$ is not continuous your statement is clearly false.
EDIT
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = begin{cases}frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x+sqrt[4]{2^{-k+1}})- sqrt{2^{-k+1}} & left(-sqrt[4]{2^{-k+1}}leq x<-sqrt[4]{2^{-k}}; kin Bbb Zright)\ 0 & (x=0)\ frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x-sqrt[4]{2^{-k+1}})+ sqrt{2^{-k+1}}& left(sqrt[4]{2^{-k}}leq x<sqrt[4]{2^{-k+1}}; kin Bbb Zright) .end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.
answered Feb 2 at 17:09
MatteoMatteo
1,142313
$endgroup$
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
add a comment |
$begingroup$
Here is a counterexample. Define
$$
f(x) = begin{cases}
0 & (x leq 0)\
-frac{1}{2^n} & left(frac{1}{sqrt{2^n}} leq x < frac{1}{sqrt{2^{n-1}}}; n > 1 right)\
-frac{1}{2} & left(xgeq frac{1}{sqrt 2}right)
end{cases}
$$
Clearly
$$f'_+(x) = lim_{hrightarrow 0^+}frac{f(x+h)-f(x)}{h}=0$$
for all $x neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 leq f(x) leq 0,$$
so that
$$-x leq frac{f(x)}{x} leq 0.$$
Thus
$$f'_+(0) = lim_{hrightarrow 0^+} frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) geq 0$, for all $xin Bbb R$, but for any $delta > 0$, $xin (0,delta) Rightarrow f(x) < 0$.
In conclusion, if $f(x)$ is not continuous your statement is clearly false.
EDIT
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = begin{cases}frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x+sqrt[4]{2^{-k+1}})- sqrt{2^{-k+1}} & left(-sqrt[4]{2^{-k+1}}leq x<-sqrt[4]{2^{-k}}; kin Bbb Zright)\ 0 & (x=0)\ frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x-sqrt[4]{2^{-k+1}})+ sqrt{2^{-k+1}}& left(sqrt[4]{2^{-k}}leq x<sqrt[4]{2^{-k+1}}; kin Bbb Zright) .end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.
answered Feb 2 at 17:09
MatteoMatteo
1,142313
$endgroup$
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
add a comment |
$begingroup$
Here is a counterexample. Define
$$
f(x) = begin{cases}
0 & (x leq 0)\
-frac{1}{2^n} & left(frac{1}{sqrt{2^n}} leq x < frac{1}{sqrt{2^{n-1}}}; n > 1 right)\
-frac{1}{2} & left(xgeq frac{1}{sqrt 2}right)
end{cases}
$$
Clearly
$$f'_+(x) = lim_{hrightarrow 0^+}frac{f(x+h)-f(x)}{h}=0$$
for all $x neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 leq f(x) leq 0,$$
so that
$$-x leq frac{f(x)}{x} leq 0.$$
Thus
$$f'_+(0) = lim_{hrightarrow 0^+} frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) geq 0$, for all $xin Bbb R$, but for any $delta > 0$, $xin (0,delta) Rightarrow f(x) < 0$.
In conclusion, if $f(x)$ is not continuous your statement is clearly false.
EDIT
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = begin{cases}frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x+sqrt[4]{2^{-k+1}})- sqrt{2^{-k+1}} & left(-sqrt[4]{2^{-k+1}}leq x<-sqrt[4]{2^{-k}}; kin Bbb Zright)\ 0 & (x=0)\ frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x-sqrt[4]{2^{-k+1}})+ sqrt{2^{-k+1}}& left(sqrt[4]{2^{-k}}leq x<sqrt[4]{2^{-k+1}}; kin Bbb Zright) .end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.
answered Feb 2 at 17:09
MatteoMatteo
1,142313
$endgroup$
Here is a counterexample. Define
$$
f(x) = begin{cases}
0 & (x leq 0)\
-frac{1}{2^n} & left(frac{1}{sqrt{2^n}} leq x < frac{1}{sqrt{2^{n-1}}}; n > 1 right)\
-frac{1}{2} & left(xgeq frac{1}{sqrt 2}right)
end{cases}
$$
Clearly
$$f'_+(x) = lim_{hrightarrow 0^+}frac{f(x+h)-f(x)}{h}=0$$
for all $x neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 leq f(x) leq 0,$$
so that
$$-x leq frac{f(x)}{x} leq 0.$$
Thus
$$f'_+(0) = lim_{hrightarrow 0^+} frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) geq 0$, for all $xin Bbb R$, but for any $delta > 0$, $xin (0,delta) Rightarrow f(x) < 0$.
In conclusion, if $f(x)$ is not continuous your statement is clearly false.
EDIT
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = begin{cases}frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x+sqrt[4]{2^{-k+1}})- sqrt{2^{-k+1}} & left(-sqrt[4]{2^{-k+1}}leq x<-sqrt[4]{2^{-k}}; kin Bbb Zright)\ 0 & (x=0)\ frac{sqrt{2^{-k}}-sqrt{2^{-k+1}}}{sqrt[4]{2^{-k+1}}-sqrt[4]{2^{-k}}}(x-sqrt[4]{2^{-k+1}})+ sqrt{2^{-k+1}}& left(sqrt[4]{2^{-k}}leq x<sqrt[4]{2^{-k+1}}; kin Bbb Zright) .end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.
answered Feb 2 at 17:09
MatteoMatteo
1,142313
edited Feb 6 at 17:49
answered Feb 2 at 17:09
MatteoMatteo
1,142313
answered Feb 2 at 17:09
MatteoMatteo
1,142313
answered Feb 2 at 17:09
MatteoMatteo
1,142313
1,142313
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
add a comment |
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
2
2
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
$begingroup$
Good counter example (+1).
$endgroup$
– Surb
Feb 2 at 20:10
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%2f3088518%2fif-the-right-derivative-is-positive-at-each-point-can-i-conclude-that-f-is-in%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
$begingroup$
Your approach doesn't. You've taken too loose an approximation. Instead, bound $|varepsilon(h)|$ by $f'_d(x)h$, and you have $f(x+h) = f(x)+f'_d(x)h + varepsilon(h) geq f(x) + f'_d(x)h - |varepsilon(h)| geq f(x) + f'_d(x)h - f'_d(x)h = f(x)$.
$endgroup$
– user3482749
Jan 26 at 17:53
$begingroup$
@user3482749: Thank you but why $|varepsilon (h)|leq f'(x)h$ ?
$endgroup$
– Dylan
Jan 26 at 17:56
$begingroup$
It's more : take $delta _x$ s.t. $|varepsilon (h)|leq f'_d(x)$ when $|h|leq delta _x$. This is possible since $varepsilon (h)to 0$
$endgroup$
– Surb
Jan 26 at 18:04
$begingroup$
@user3482749, I was curious about the question and read your comment. I didnt' get exactly your steps. My doubt is that you apparently (but I may be wrong) did not use the hypothesis that $f'_d(x)$ is non negative everywhere. This is definitely necessary for the statement to hold, If $f'_d(x)$ is non negative only at, say, $x_0$ then no conclusion can be drawn on the behavior of $f(x)$ in a neighborhood of $x_0$. Am I right?
$endgroup$
– Matteo
Jan 26 at 20:39
$begingroup$
It's not "derivable". It's "differentiable".
$endgroup$
– MathematicsStudent1122
Feb 2 at 17:24