Mixing
Have I correctly proved that $lim_{||(x,y)||toinfty}frac1{y-x}int_x^yexp(-1/|t|)dt$ equals $1$?
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I should prove that, as long as $yne x$, $$f(x,y)=frac1{y-x}int_x^yexp(-1/|t|)dtlongrightarrow1 text{as $||(x,y)||toinfty$}$$ and I would like to do it without $varepsilon-delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|toinfty$ and $|h|nottoinfty$, as $||(x,y)||toinfty$.
We have $$f(x,y)=f_h(x)=frac1{h}int_x^{x+h}exp(-1/|t|)dt=frac1hint_0^hexp(-1/|t+x|)dt.$$ If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$.
In the former case, by the dominated convergence theorem, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=1;$$in the latter case, by De L'Hospital, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}exp(-1/|h+x|)=lim_{|y|toinfty}exp(-1/|y|)=1. $$
Finally, suppose $|h|nottoinfty$; then $|x|toinfty$, as a consequence of begin{align}sqrt{x^2+y^2}=sqrt{x^2+(x+h)^2}=sqrt{2x^2+2hx+h^2}&lesqrt{4x^2+4|hx|+h^2} \ &lesqrt{(2|x|+|h|)^2} \ &=2|x|+|h|.end{align}So we get once again$$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=lim_{||(x,y)||toinfty}frac{h}h=1.$$
Is my proof correct? Otherwise, how can I amend it?
real-analysis integration multivariable-calculus proof-verification alternative-proof
asked Jan 9 at 10:14
LearnerLearner
17510
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show 1 more comment
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I should prove that, as long as $yne x$, $$f(x,y)=frac1{y-x}int_x^yexp(-1/|t|)dtlongrightarrow1 text{as $||(x,y)||toinfty$}$$ and I would like to do it without $varepsilon-delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|toinfty$ and $|h|nottoinfty$, as $||(x,y)||toinfty$.
We have $$f(x,y)=f_h(x)=frac1{h}int_x^{x+h}exp(-1/|t|)dt=frac1hint_0^hexp(-1/|t+x|)dt.$$ If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$.
In the former case, by the dominated convergence theorem, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=1;$$in the latter case, by De L'Hospital, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}exp(-1/|h+x|)=lim_{|y|toinfty}exp(-1/|y|)=1. $$
Finally, suppose $|h|nottoinfty$; then $|x|toinfty$, as a consequence of begin{align}sqrt{x^2+y^2}=sqrt{x^2+(x+h)^2}=sqrt{2x^2+2hx+h^2}&lesqrt{4x^2+4|hx|+h^2} \ &lesqrt{(2|x|+|h|)^2} \ &=2|x|+|h|.end{align}So we get once again$$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=lim_{||(x,y)||toinfty}frac{h}h=1.$$
Is my proof correct? Otherwise, how can I amend it?
real-analysis integration multivariable-calculus proof-verification alternative-proof
asked Jan 9 at 10:14
LearnerLearner
17510
$endgroup$
$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
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– mathworker21
Jan 12 at 17:56
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@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
$endgroup$
– Learner
Jan 12 at 18:15
$begingroup$
you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
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– mathworker21
Jan 12 at 18:39
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
$endgroup$
– Learner
Jan 12 at 21:56
1
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
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– zhw.
Jan 14 at 18:00
|
show 1 more comment
$begingroup$
I should prove that, as long as $yne x$, $$f(x,y)=frac1{y-x}int_x^yexp(-1/|t|)dtlongrightarrow1 text{as $||(x,y)||toinfty$}$$ and I would like to do it without $varepsilon-delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|toinfty$ and $|h|nottoinfty$, as $||(x,y)||toinfty$.
We have $$f(x,y)=f_h(x)=frac1{h}int_x^{x+h}exp(-1/|t|)dt=frac1hint_0^hexp(-1/|t+x|)dt.$$ If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$.
In the former case, by the dominated convergence theorem, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=1;$$in the latter case, by De L'Hospital, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}exp(-1/|h+x|)=lim_{|y|toinfty}exp(-1/|y|)=1. $$
Finally, suppose $|h|nottoinfty$; then $|x|toinfty$, as a consequence of begin{align}sqrt{x^2+y^2}=sqrt{x^2+(x+h)^2}=sqrt{2x^2+2hx+h^2}&lesqrt{4x^2+4|hx|+h^2} \ &lesqrt{(2|x|+|h|)^2} \ &=2|x|+|h|.end{align}So we get once again$$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=lim_{||(x,y)||toinfty}frac{h}h=1.$$
Is my proof correct? Otherwise, how can I amend it?
real-analysis integration multivariable-calculus proof-verification alternative-proof
asked Jan 9 at 10:14
LearnerLearner
17510
$endgroup$
I should prove that, as long as $yne x$, $$f(x,y)=frac1{y-x}int_x^yexp(-1/|t|)dtlongrightarrow1 text{as $||(x,y)||toinfty$}$$ and I would like to do it without $varepsilon-delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|toinfty$ and $|h|nottoinfty$, as $||(x,y)||toinfty$.
We have $$f(x,y)=f_h(x)=frac1{h}int_x^{x+h}exp(-1/|t|)dt=frac1hint_0^hexp(-1/|t+x|)dt.$$ If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$.
In the former case, by the dominated convergence theorem, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=1;$$in the latter case, by De L'Hospital, $$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}exp(-1/|h+x|)=lim_{|y|toinfty}exp(-1/|y|)=1. $$
Finally, suppose $|h|nottoinfty$; then $|x|toinfty$, as a consequence of begin{align}sqrt{x^2+y^2}=sqrt{x^2+(x+h)^2}=sqrt{2x^2+2hx+h^2}&lesqrt{4x^2+4|hx|+h^2} \ &lesqrt{(2|x|+|h|)^2} \ &=2|x|+|h|.end{align}So we get once again$$lim_{||(x,y)||toinfty}frac1hint_0^hexp(-1/|t+x|)dt=lim_{||(x,y)||toinfty}frac1hint_0^hdt=lim_{||(x,y)||toinfty}frac{h}h=1.$$
Is my proof correct? Otherwise, how can I amend it?
real-analysis integration multivariable-calculus proof-verification alternative-proof
real-analysis integration multivariable-calculus proof-verification alternative-proof
asked Jan 9 at 10:14
LearnerLearner
17510
asked Jan 9 at 10:14
LearnerLearner
17510
edited Jan 12 at 18:16
Learner
asked Jan 9 at 10:14
LearnerLearner
17510
asked Jan 9 at 10:14
LearnerLearner
17510
asked Jan 9 at 10:14
LearnerLearner
17510
17510
$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
$endgroup$
– mathworker21
Jan 12 at 17:56
$begingroup$
@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
$endgroup$
– Learner
Jan 12 at 18:15
$begingroup$
you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
$endgroup$
– mathworker21
Jan 12 at 18:39
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
$endgroup$
– Learner
Jan 12 at 21:56
1
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
$endgroup$
– zhw.
Jan 14 at 18:00
|
show 1 more comment
$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
$endgroup$
– mathworker21
Jan 12 at 17:56
$begingroup$
@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
$endgroup$
– Learner
Jan 12 at 18:15
$begingroup$
you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
$endgroup$
– mathworker21
Jan 12 at 18:39
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
$endgroup$
– Learner
Jan 12 at 21:56
1
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
$endgroup$
– zhw.
Jan 14 at 18:00
$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
$endgroup$
– mathworker21
Jan 12 at 17:56
$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
$endgroup$
– mathworker21
Jan 12 at 17:56
$begingroup$
@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
$endgroup$
– Learner
Jan 12 at 18:15
$begingroup$
@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
$endgroup$
– Learner
Jan 12 at 18:15
$begingroup$
you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
$endgroup$
– mathworker21
Jan 12 at 18:39
$begingroup$
you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
$endgroup$
– mathworker21
Jan 12 at 18:39
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
$endgroup$
– Learner
Jan 12 at 21:56
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
$endgroup$
– Learner
Jan 12 at 21:56
1
1
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
$endgroup$
– zhw.
Jan 14 at 18:00
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
$endgroup$
– zhw.
Jan 14 at 18:00
|
show 1 more comment
1 Answer
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As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:mathbb Rto mathbb R$ is continuous and $lim_{|x|to infty} f(x) = Lin mathbb R.$ For $xne y,$ define
$$A(x,y) = frac{1}{y-x}int_x^y f.$$
Then $lim_{|(x,y)|to infty} A(x,y) = L.$
One special case is easy: $lim_{|y|to infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|to infty$ such that $A(x_n,y_n)to L'ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_nto infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$tag 1 A(x_n,y_n) = int_0^1 f(x_n+t(y_n -x_n)),dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_nto L$ pointwise on $ (0,1].$ By the DCT, $(1)to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_nto infty.$ Then $x_n+t(y_n -x_n)to infty$ for each $tin [0,1].$ Again using DCT, we see $(1)to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_nto -infty.$ For large $n,$ we have
$$tag 2 A(x_n,y_n) = frac{-x_n}{y_n-x_n}A(x_n,0) + frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_nto L,$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n to L.$ This shows $(2)to L,$ and again we have a contradiction.
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
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add a comment |
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$begingroup$
As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:mathbb Rto mathbb R$ is continuous and $lim_{|x|to infty} f(x) = Lin mathbb R.$ For $xne y,$ define
$$A(x,y) = frac{1}{y-x}int_x^y f.$$
Then $lim_{|(x,y)|to infty} A(x,y) = L.$
One special case is easy: $lim_{|y|to infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|to infty$ such that $A(x_n,y_n)to L'ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_nto infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$tag 1 A(x_n,y_n) = int_0^1 f(x_n+t(y_n -x_n)),dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_nto L$ pointwise on $ (0,1].$ By the DCT, $(1)to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_nto infty.$ Then $x_n+t(y_n -x_n)to infty$ for each $tin [0,1].$ Again using DCT, we see $(1)to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_nto -infty.$ For large $n,$ we have
$$tag 2 A(x_n,y_n) = frac{-x_n}{y_n-x_n}A(x_n,0) + frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_nto L,$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n to L.$ This shows $(2)to L,$ and again we have a contradiction.
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
$endgroup$
add a comment |
$begingroup$
As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:mathbb Rto mathbb R$ is continuous and $lim_{|x|to infty} f(x) = Lin mathbb R.$ For $xne y,$ define
$$A(x,y) = frac{1}{y-x}int_x^y f.$$
Then $lim_{|(x,y)|to infty} A(x,y) = L.$
One special case is easy: $lim_{|y|to infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|to infty$ such that $A(x_n,y_n)to L'ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_nto infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$tag 1 A(x_n,y_n) = int_0^1 f(x_n+t(y_n -x_n)),dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_nto L$ pointwise on $ (0,1].$ By the DCT, $(1)to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_nto infty.$ Then $x_n+t(y_n -x_n)to infty$ for each $tin [0,1].$ Again using DCT, we see $(1)to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_nto -infty.$ For large $n,$ we have
$$tag 2 A(x_n,y_n) = frac{-x_n}{y_n-x_n}A(x_n,0) + frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_nto L,$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n to L.$ This shows $(2)to L,$ and again we have a contradiction.
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
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$begingroup$
As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:mathbb Rto mathbb R$ is continuous and $lim_{|x|to infty} f(x) = Lin mathbb R.$ For $xne y,$ define
$$A(x,y) = frac{1}{y-x}int_x^y f.$$
Then $lim_{|(x,y)|to infty} A(x,y) = L.$
One special case is easy: $lim_{|y|to infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|to infty$ such that $A(x_n,y_n)to L'ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_nto infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$tag 1 A(x_n,y_n) = int_0^1 f(x_n+t(y_n -x_n)),dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_nto L$ pointwise on $ (0,1].$ By the DCT, $(1)to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_nto infty.$ Then $x_n+t(y_n -x_n)to infty$ for each $tin [0,1].$ Again using DCT, we see $(1)to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_nto -infty.$ For large $n,$ we have
$$tag 2 A(x_n,y_n) = frac{-x_n}{y_n-x_n}A(x_n,0) + frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_nto L,$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n to L.$ This shows $(2)to L,$ and again we have a contradiction.
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
$endgroup$
As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:mathbb Rto mathbb R$ is continuous and $lim_{|x|to infty} f(x) = Lin mathbb R.$ For $xne y,$ define
$$A(x,y) = frac{1}{y-x}int_x^y f.$$
Then $lim_{|(x,y)|to infty} A(x,y) = L.$
One special case is easy: $lim_{|y|to infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|to infty$ such that $A(x_n,y_n)to L'ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_nto infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$tag 1 A(x_n,y_n) = int_0^1 f(x_n+t(y_n -x_n)),dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_nto L$ pointwise on $ (0,1].$ By the DCT, $(1)to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_nto infty.$ Then $x_n+t(y_n -x_n)to infty$ for each $tin [0,1].$ Again using DCT, we see $(1)to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_nto -infty.$ For large $n,$ we have
$$tag 2 A(x_n,y_n) = frac{-x_n}{y_n-x_n}A(x_n,0) + frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_nto L,$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n to L.$ This shows $(2)to L,$ and again we have a contradiction.
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
edited Jan 17 at 19:30
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
answered Jan 16 at 19:31
zhw.zhw.
72.6k43175
72.6k43175
add a comment |
add a comment |
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$begingroup$
It's very unclear how you're using dominated convergence theorem. $h$ depends on $x$.. it seem's you're first taking a limit as $x to infty$ and then a limit as $h to infty$??
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– mathworker21
Jan 12 at 17:56
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@mathworker21 Perhaps I did it incorrectly, but it's always $||(x,y)||$ that goes to $infty$ - however in that instance I was momentarily considering the case where, also, both $|h|$ and $|x|$ do. In particular I somewhat apply DCT with $x$, and then what remains is actually identically equal to $1$. Let me remove that confusing $|h|toinfty $, and thanks for your input. What do you think?
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– Learner
Jan 12 at 18:15
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you still have $lim_{||(x,y)|| to infty}$ there but got rid of the $x$. If you want to end up with a rigorous proof, be explicit about these limits
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– mathworker21
Jan 12 at 18:39
$begingroup$
@mathworker21 What do you mean? In general it's not strange to have something like $lim_{xtoinfty}a(z)=a(z) $ where $a (z)$ is constant with respect to $x$
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– Learner
Jan 12 at 21:56
1
$begingroup$
"If $|h|toinfty$ then at least one of $|x|$ and $|y|$ must also go to $infty$." That's not true. Consider $x_n =0,1,0,2,0,3,dots,$ $y_n = 1,0,2,0,3,0,dots.$ Then $y_n-x_n = 1,-1,2,-2,3,-3,dots.$ Thus $|y_n-x_n|to infty,$ but neither $x_n$ or $y_n$ does so.
$endgroup$
– zhw.
Jan 14 at 18:00