Proof of equicontinuity via uniform convergence
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I am studying for an entrance exam in August and I am trying to review and cover the real analysis section. I came across this problem on one of the past exams and I am having some trouble with it.
Let $G(x, y)$ be a continuous function on $R^2$ and suppose for each positive
integer $k$, that $g_k$ is a continuous function defined on [0, 1] with the property that $|g_k(y)| leq 1$ for all $y in [0, 1]$. Now define
$$f_k(x) := int_{0}^{1}g_k(y)G(x, y) dy.$$
Prove that the sequence ${f_k}$ is equicontinuous on [0, 1].
So for the problem above, given that $[0,1]$ is compact, I think that it is sufficient to show that {${f_k(x)}$} is uniformly convergent on the interval $[0,1]$. How would I show that {${f_k(x)}$} is uniformly convergent on {${f_k(x)}$}?
real-analysis uniform-convergence equicontinuity
asked Jan 21 at 21:42
geoplantedgeoplanted
597
$endgroup$
add a comment |
$begingroup$
I am studying for an entrance exam in August and I am trying to review and cover the real analysis section. I came across this problem on one of the past exams and I am having some trouble with it.
Let $G(x, y)$ be a continuous function on $R^2$ and suppose for each positive
integer $k$, that $g_k$ is a continuous function defined on [0, 1] with the property that $|g_k(y)| leq 1$ for all $y in [0, 1]$. Now define
$$f_k(x) := int_{0}^{1}g_k(y)G(x, y) dy.$$
Prove that the sequence ${f_k}$ is equicontinuous on [0, 1].
So for the problem above, given that $[0,1]$ is compact, I think that it is sufficient to show that {${f_k(x)}$} is uniformly convergent on the interval $[0,1]$. How would I show that {${f_k(x)}$} is uniformly convergent on {${f_k(x)}$}?
real-analysis uniform-convergence equicontinuity
asked Jan 21 at 21:42
geoplantedgeoplanted
597
$endgroup$
1
$begingroup$
Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
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Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23
add a comment |
$begingroup$
I am studying for an entrance exam in August and I am trying to review and cover the real analysis section. I came across this problem on one of the past exams and I am having some trouble with it.
Let $G(x, y)$ be a continuous function on $R^2$ and suppose for each positive
integer $k$, that $g_k$ is a continuous function defined on [0, 1] with the property that $|g_k(y)| leq 1$ for all $y in [0, 1]$. Now define
$$f_k(x) := int_{0}^{1}g_k(y)G(x, y) dy.$$
Prove that the sequence ${f_k}$ is equicontinuous on [0, 1].
So for the problem above, given that $[0,1]$ is compact, I think that it is sufficient to show that {${f_k(x)}$} is uniformly convergent on the interval $[0,1]$. How would I show that {${f_k(x)}$} is uniformly convergent on {${f_k(x)}$}?
real-analysis uniform-convergence equicontinuity
asked Jan 21 at 21:42
geoplantedgeoplanted
597
$endgroup$
I am studying for an entrance exam in August and I am trying to review and cover the real analysis section. I came across this problem on one of the past exams and I am having some trouble with it.
Let $G(x, y)$ be a continuous function on $R^2$ and suppose for each positive
integer $k$, that $g_k$ is a continuous function defined on [0, 1] with the property that $|g_k(y)| leq 1$ for all $y in [0, 1]$. Now define
$$f_k(x) := int_{0}^{1}g_k(y)G(x, y) dy.$$
Prove that the sequence ${f_k}$ is equicontinuous on [0, 1].
So for the problem above, given that $[0,1]$ is compact, I think that it is sufficient to show that {${f_k(x)}$} is uniformly convergent on the interval $[0,1]$. How would I show that {${f_k(x)}$} is uniformly convergent on {${f_k(x)}$}?
real-analysis uniform-convergence equicontinuity
real-analysis uniform-convergence equicontinuity
asked Jan 21 at 21:42
geoplantedgeoplanted
597
asked Jan 21 at 21:42
geoplantedgeoplanted
597
edited Jan 21 at 23:21
geoplanted
asked Jan 21 at 21:42
geoplantedgeoplanted
597
asked Jan 21 at 21:42
geoplantedgeoplanted
597
asked Jan 21 at 21:42
geoplantedgeoplanted
597
597
1
$begingroup$
Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
$begingroup$
Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23
add a comment |
1
$begingroup$
Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
$begingroup$
Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23
1
1
$begingroup$
Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
$begingroup$
Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
$begingroup$
Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23
$begingroup$
Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23
add a comment |
1 Answer
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You cannot prove convergence. Instead, take $epsilon >0$ and choose $delta >0$ such that $|G(x,y)-G(x',y')| <epsilon$ whenever $|(x,y)-(x',y')| <delta$. This is possible because any continuous function on the compact space $[0,1]times [0,1]$ is unifomly continuous. From this you get $|f_k(x)-f_k(y)| <epsilon$ whenever $|x-y| <delta$ which is the definition of equicontinuity.
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
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add a comment |
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You cannot prove convergence. Instead, take $epsilon >0$ and choose $delta >0$ such that $|G(x,y)-G(x',y')| <epsilon$ whenever $|(x,y)-(x',y')| <delta$. This is possible because any continuous function on the compact space $[0,1]times [0,1]$ is unifomly continuous. From this you get $|f_k(x)-f_k(y)| <epsilon$ whenever $|x-y| <delta$ which is the definition of equicontinuity.
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
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add a comment |
$begingroup$
You cannot prove convergence. Instead, take $epsilon >0$ and choose $delta >0$ such that $|G(x,y)-G(x',y')| <epsilon$ whenever $|(x,y)-(x',y')| <delta$. This is possible because any continuous function on the compact space $[0,1]times [0,1]$ is unifomly continuous. From this you get $|f_k(x)-f_k(y)| <epsilon$ whenever $|x-y| <delta$ which is the definition of equicontinuity.
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
You cannot prove convergence. Instead, take $epsilon >0$ and choose $delta >0$ such that $|G(x,y)-G(x',y')| <epsilon$ whenever $|(x,y)-(x',y')| <delta$. This is possible because any continuous function on the compact space $[0,1]times [0,1]$ is unifomly continuous. From this you get $|f_k(x)-f_k(y)| <epsilon$ whenever $|x-y| <delta$ which is the definition of equicontinuity.
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
You cannot prove convergence. Instead, take $epsilon >0$ and choose $delta >0$ such that $|G(x,y)-G(x',y')| <epsilon$ whenever $|(x,y)-(x',y')| <delta$. This is possible because any continuous function on the compact space $[0,1]times [0,1]$ is unifomly continuous. From this you get $|f_k(x)-f_k(y)| <epsilon$ whenever $|x-y| <delta$ which is the definition of equicontinuity.
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:58
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
add a comment |
add a comment |
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Hint: prove that for each $x$, and $epsilon >0$ there is a small interval $I_x$ around $x$ such that for all $x’ in I_x$, $0 leq y leq 1$, $|G(x’,y)-G(x,y)| leq epsilon$. Prive that such an $I_x$ is an “interval of $epsilon$-continuity” for each $f_k$.
$endgroup$
– Mindlack
Jan 21 at 21:56
$begingroup$
Whoops, I goofed. I accidentally wrote "uniformly continuous" instead of "uniformly convergent" in the last paragraph. My mistake, I fixed it. So do your comments still apply?
$endgroup$
– geoplanted
Jan 21 at 23:23