Mixing
Showing an integral operator is compact
$begingroup$
Let $Kin L^2(mathbb{R}^2)$, and define the operator $T: L^2(mathbb{R})to L^2(mathbb{R})$ by
$$(Tf)(x) = int K(x, y) f(y) dy$$
I want to show this operator is compact (I imagine this is a standard result).
I found a proof (I imagine the standard one) of this using that this operator is the limit of finite rank operators. I had another idea though, and I was wondering if it could be fleshed out:
Let $B$ be a collection of functions in $L^2$ with norm $leq 1$. The function $G(x) = lvert lvert K(x, cdot) rvertrvert_2$ is in $L^2$, and for all $fin B$, we have $lvert Tfrvert leq G$. So all functions in $T(B)$, which we want to have compact closure, fit below this one $L^2$ function.
Given a sequence of functions in $T(B)$, if we approximated them in $L^2$ with continuous functions and then found a uniform-on-compact-sets limit of these functions, we could prove the theorem. But for this of course, per Arzela-Ascoli, we need equicontinuity.
Towards proving this, let $L_a: L^2 to L^2$ be the shift operator $fmapsto fcirc (xmapsto x + a)$. We can show that
$$lvert lvert Tf - L_acirc Tfrvert rvert_2 to 0$$
for all $fin B$ uniformly. This certainly smacks of equicontinuity.
My question is: from this above assumption, that all functions in $T(B)$ have this "$L^2$ equicontinuity", can we find continuous functions approximating them to arbitrary precision which are also equicontinuous?
real-analysis integration functional-analysis
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
Let $Kin L^2(mathbb{R}^2)$, and define the operator $T: L^2(mathbb{R})to L^2(mathbb{R})$ by
$$(Tf)(x) = int K(x, y) f(y) dy$$
I want to show this operator is compact (I imagine this is a standard result).
I found a proof (I imagine the standard one) of this using that this operator is the limit of finite rank operators. I had another idea though, and I was wondering if it could be fleshed out:
Let $B$ be a collection of functions in $L^2$ with norm $leq 1$. The function $G(x) = lvert lvert K(x, cdot) rvertrvert_2$ is in $L^2$, and for all $fin B$, we have $lvert Tfrvert leq G$. So all functions in $T(B)$, which we want to have compact closure, fit below this one $L^2$ function.
Given a sequence of functions in $T(B)$, if we approximated them in $L^2$ with continuous functions and then found a uniform-on-compact-sets limit of these functions, we could prove the theorem. But for this of course, per Arzela-Ascoli, we need equicontinuity.
Towards proving this, let $L_a: L^2 to L^2$ be the shift operator $fmapsto fcirc (xmapsto x + a)$. We can show that
$$lvert lvert Tf - L_acirc Tfrvert rvert_2 to 0$$
for all $fin B$ uniformly. This certainly smacks of equicontinuity.
My question is: from this above assumption, that all functions in $T(B)$ have this "$L^2$ equicontinuity", can we find continuous functions approximating them to arbitrary precision which are also equicontinuous?
real-analysis integration functional-analysis
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
$endgroup$
$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13
add a comment |
$begingroup$
Let $Kin L^2(mathbb{R}^2)$, and define the operator $T: L^2(mathbb{R})to L^2(mathbb{R})$ by
$$(Tf)(x) = int K(x, y) f(y) dy$$
I want to show this operator is compact (I imagine this is a standard result).
I found a proof (I imagine the standard one) of this using that this operator is the limit of finite rank operators. I had another idea though, and I was wondering if it could be fleshed out:
Let $B$ be a collection of functions in $L^2$ with norm $leq 1$. The function $G(x) = lvert lvert K(x, cdot) rvertrvert_2$ is in $L^2$, and for all $fin B$, we have $lvert Tfrvert leq G$. So all functions in $T(B)$, which we want to have compact closure, fit below this one $L^2$ function.
Given a sequence of functions in $T(B)$, if we approximated them in $L^2$ with continuous functions and then found a uniform-on-compact-sets limit of these functions, we could prove the theorem. But for this of course, per Arzela-Ascoli, we need equicontinuity.
Towards proving this, let $L_a: L^2 to L^2$ be the shift operator $fmapsto fcirc (xmapsto x + a)$. We can show that
$$lvert lvert Tf - L_acirc Tfrvert rvert_2 to 0$$
for all $fin B$ uniformly. This certainly smacks of equicontinuity.
My question is: from this above assumption, that all functions in $T(B)$ have this "$L^2$ equicontinuity", can we find continuous functions approximating them to arbitrary precision which are also equicontinuous?
real-analysis integration functional-analysis
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
$endgroup$
Let $Kin L^2(mathbb{R}^2)$, and define the operator $T: L^2(mathbb{R})to L^2(mathbb{R})$ by
$$(Tf)(x) = int K(x, y) f(y) dy$$
I want to show this operator is compact (I imagine this is a standard result).
I found a proof (I imagine the standard one) of this using that this operator is the limit of finite rank operators. I had another idea though, and I was wondering if it could be fleshed out:
Let $B$ be a collection of functions in $L^2$ with norm $leq 1$. The function $G(x) = lvert lvert K(x, cdot) rvertrvert_2$ is in $L^2$, and for all $fin B$, we have $lvert Tfrvert leq G$. So all functions in $T(B)$, which we want to have compact closure, fit below this one $L^2$ function.
Given a sequence of functions in $T(B)$, if we approximated them in $L^2$ with continuous functions and then found a uniform-on-compact-sets limit of these functions, we could prove the theorem. But for this of course, per Arzela-Ascoli, we need equicontinuity.
Towards proving this, let $L_a: L^2 to L^2$ be the shift operator $fmapsto fcirc (xmapsto x + a)$. We can show that
$$lvert lvert Tf - L_acirc Tfrvert rvert_2 to 0$$
for all $fin B$ uniformly. This certainly smacks of equicontinuity.
My question is: from this above assumption, that all functions in $T(B)$ have this "$L^2$ equicontinuity", can we find continuous functions approximating them to arbitrary precision which are also equicontinuous?
real-analysis integration functional-analysis
real-analysis integration functional-analysis
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
asked Jan 24 at 1:27
Van LatimerVan Latimer
363110
363110
$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13
add a comment |
$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13
$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13
add a comment |
1 Answer
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I think I've got it. For each $fin T(B)$, let $f_a(x) = frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set ${f_a: fin T(B)}$ is equicontinuous, because by Cauchy-Schwarz,
$$lvert f_a(y) - f_a(z) leq frac{1}{a}int lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert lvert f(x) rvert dx$$
$$leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert frvert rvert_2 leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert Grvert rvert_2$$
which goes to 0 uniformly as $|x- y|to 0$ for all $f$.
Now see that $f_a(x) to f$ in $L^2$ as $ato 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(int_A f dmu)^2 leq mu(A) int f^2$), and then Tonelli's theorem,
$$lvert lvert f_a - frvert rvert_2^2 = int lvert left(frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dyright) - f(x)rvert^2dx = intlvert frac{int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} rvert^2dx$$
$$leq intfrac{aint_{x-1/2}^{x+a/2} lvert f(y) - f(x)rvert^2dy}{a^2} dx =frac{1}{a} int int 1_{{x-a/2 < y < x + a/2}} |f(x) - f(y)|^2dxdy$$
$$ = frac{1}{a} int_0^a int |f(x + delta) - f(x)|^2dx ddelta$$
If now $a$ is small enough that $lvertlvert f - L_deltacirc frvert rvert_2 < epsilon$ whenever $delta < a$, we have that $lvert lvert f_a - frvertrvert_2^2 < epsilon$.
These things being established, let's define a sequence in a metric space to be $epsilon$-Cauchy if eventually its terms differ by less than $epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous.
Now, given a sequence $f_n in T(B)$, let let $a$ be such that $lvertlvert(f_n)_a-f_nrvertrvert_2 < epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2epsilon$-Cauchy.
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $epsilon$-Cauchy for every $epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$.
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
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$begingroup$
I think I've got it. For each $fin T(B)$, let $f_a(x) = frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set ${f_a: fin T(B)}$ is equicontinuous, because by Cauchy-Schwarz,
$$lvert f_a(y) - f_a(z) leq frac{1}{a}int lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert lvert f(x) rvert dx$$
$$leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert frvert rvert_2 leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert Grvert rvert_2$$
which goes to 0 uniformly as $|x- y|to 0$ for all $f$.
Now see that $f_a(x) to f$ in $L^2$ as $ato 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(int_A f dmu)^2 leq mu(A) int f^2$), and then Tonelli's theorem,
$$lvert lvert f_a - frvert rvert_2^2 = int lvert left(frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dyright) - f(x)rvert^2dx = intlvert frac{int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} rvert^2dx$$
$$leq intfrac{aint_{x-1/2}^{x+a/2} lvert f(y) - f(x)rvert^2dy}{a^2} dx =frac{1}{a} int int 1_{{x-a/2 < y < x + a/2}} |f(x) - f(y)|^2dxdy$$
$$ = frac{1}{a} int_0^a int |f(x + delta) - f(x)|^2dx ddelta$$
If now $a$ is small enough that $lvertlvert f - L_deltacirc frvert rvert_2 < epsilon$ whenever $delta < a$, we have that $lvert lvert f_a - frvertrvert_2^2 < epsilon$.
These things being established, let's define a sequence in a metric space to be $epsilon$-Cauchy if eventually its terms differ by less than $epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous.
Now, given a sequence $f_n in T(B)$, let let $a$ be such that $lvertlvert(f_n)_a-f_nrvertrvert_2 < epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2epsilon$-Cauchy.
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $epsilon$-Cauchy for every $epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$.
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
I think I've got it. For each $fin T(B)$, let $f_a(x) = frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set ${f_a: fin T(B)}$ is equicontinuous, because by Cauchy-Schwarz,
$$lvert f_a(y) - f_a(z) leq frac{1}{a}int lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert lvert f(x) rvert dx$$
$$leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert frvert rvert_2 leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert Grvert rvert_2$$
which goes to 0 uniformly as $|x- y|to 0$ for all $f$.
Now see that $f_a(x) to f$ in $L^2$ as $ato 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(int_A f dmu)^2 leq mu(A) int f^2$), and then Tonelli's theorem,
$$lvert lvert f_a - frvert rvert_2^2 = int lvert left(frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dyright) - f(x)rvert^2dx = intlvert frac{int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} rvert^2dx$$
$$leq intfrac{aint_{x-1/2}^{x+a/2} lvert f(y) - f(x)rvert^2dy}{a^2} dx =frac{1}{a} int int 1_{{x-a/2 < y < x + a/2}} |f(x) - f(y)|^2dxdy$$
$$ = frac{1}{a} int_0^a int |f(x + delta) - f(x)|^2dx ddelta$$
If now $a$ is small enough that $lvertlvert f - L_deltacirc frvert rvert_2 < epsilon$ whenever $delta < a$, we have that $lvert lvert f_a - frvertrvert_2^2 < epsilon$.
These things being established, let's define a sequence in a metric space to be $epsilon$-Cauchy if eventually its terms differ by less than $epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous.
Now, given a sequence $f_n in T(B)$, let let $a$ be such that $lvertlvert(f_n)_a-f_nrvertrvert_2 < epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2epsilon$-Cauchy.
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $epsilon$-Cauchy for every $epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$.
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
I think I've got it. For each $fin T(B)$, let $f_a(x) = frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set ${f_a: fin T(B)}$ is equicontinuous, because by Cauchy-Schwarz,
$$lvert f_a(y) - f_a(z) leq frac{1}{a}int lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert lvert f(x) rvert dx$$
$$leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert frvert rvert_2 leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert Grvert rvert_2$$
which goes to 0 uniformly as $|x- y|to 0$ for all $f$.
Now see that $f_a(x) to f$ in $L^2$ as $ato 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(int_A f dmu)^2 leq mu(A) int f^2$), and then Tonelli's theorem,
$$lvert lvert f_a - frvert rvert_2^2 = int lvert left(frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dyright) - f(x)rvert^2dx = intlvert frac{int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} rvert^2dx$$
$$leq intfrac{aint_{x-1/2}^{x+a/2} lvert f(y) - f(x)rvert^2dy}{a^2} dx =frac{1}{a} int int 1_{{x-a/2 < y < x + a/2}} |f(x) - f(y)|^2dxdy$$
$$ = frac{1}{a} int_0^a int |f(x + delta) - f(x)|^2dx ddelta$$
If now $a$ is small enough that $lvertlvert f - L_deltacirc frvert rvert_2 < epsilon$ whenever $delta < a$, we have that $lvert lvert f_a - frvertrvert_2^2 < epsilon$.
These things being established, let's define a sequence in a metric space to be $epsilon$-Cauchy if eventually its terms differ by less than $epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous.
Now, given a sequence $f_n in T(B)$, let let $a$ be such that $lvertlvert(f_n)_a-f_nrvertrvert_2 < epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2epsilon$-Cauchy.
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $epsilon$-Cauchy for every $epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$.
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
$endgroup$
I think I've got it. For each $fin T(B)$, let $f_a(x) = frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set ${f_a: fin T(B)}$ is equicontinuous, because by Cauchy-Schwarz,
$$lvert f_a(y) - f_a(z) leq frac{1}{a}int lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert lvert f(x) rvert dx$$
$$leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert frvert rvert_2 leq frac{1}{a} lvert lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} rvert rvert_2 lvert lvert Grvert rvert_2$$
which goes to 0 uniformly as $|x- y|to 0$ for all $f$.
Now see that $f_a(x) to f$ in $L^2$ as $ato 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(int_A f dmu)^2 leq mu(A) int f^2$), and then Tonelli's theorem,
$$lvert lvert f_a - frvert rvert_2^2 = int lvert left(frac{1}{a}int_{x-a/2}^{x + a/2} f(y)dyright) - f(x)rvert^2dx = intlvert frac{int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} rvert^2dx$$
$$leq intfrac{aint_{x-1/2}^{x+a/2} lvert f(y) - f(x)rvert^2dy}{a^2} dx =frac{1}{a} int int 1_{{x-a/2 < y < x + a/2}} |f(x) - f(y)|^2dxdy$$
$$ = frac{1}{a} int_0^a int |f(x + delta) - f(x)|^2dx ddelta$$
If now $a$ is small enough that $lvertlvert f - L_deltacirc frvert rvert_2 < epsilon$ whenever $delta < a$, we have that $lvert lvert f_a - frvertrvert_2^2 < epsilon$.
These things being established, let's define a sequence in a metric space to be $epsilon$-Cauchy if eventually its terms differ by less than $epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous.
Now, given a sequence $f_n in T(B)$, let let $a$ be such that $lvertlvert(f_n)_a-f_nrvertrvert_2 < epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2epsilon$-Cauchy.
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $epsilon$-Cauchy for every $epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$.
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
answered Jan 28 at 14:49
Van LatimerVan Latimer
363110
363110
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$begingroup$
You are using several steps of approximation, first approximating $Tf_n$ by some continuous functions $g_n$, and then $Tf_n$ by $L_a Tf_n$. But it certainly does not work: in an attempt of applying Arzela-Ascoli theorem, you should show $|g_n-L_acirc g_n|_{L^infty(K)}<epsilon$ for all $nge 1$ and for all $|a|<delta$ on every compact $K$. Your method cannot give this uniform bound. In fact, the result is closely related to the notion of Hilbert-Schmidt operator. You might look for it.
$endgroup$
– Song
Jan 24 at 6:10
$begingroup$
@Song Thanks for your help. Of course we can't get continuous functions approximating even one function in L^2 to remain equicontinuous, but I think something still works, which I'll write up soon if I can
$endgroup$
– Van Latimer
Jan 24 at 15:13