Hilbert-Schmidt integral operator is compact.
In what follows, we will prove that every Hilbert-Schmidt integral operator is compact.
Formally, let k\in L_2([0,1]\times[0,1]) and T:L_2[0,1]\to L_2[0,1] be defined by
Tf(x):=\int_0^1k(x,y)f(y)dy.
Then T is compact operator.
Although the proof may seem a bit long at first sight, it is proceeds naturally, utilizing approximation of k with continuous functions.
\textbf{Proof}. We approximate k with a sequence of continuous functions, then prove that each of the corresponding operators has image contained in C[0,1], then show that each of these operators is compact when C[0,1] is equipped with the \sup norm. This will be done via the Arzela-Ascoli theorem. Finally show that these converge uniformly (in the operator norm) to the original operator, and since they are compact, the original must be as well.
The Cauchy-Bunyakovsky-Schwartz inequality is used throughout the proof.
Thus we start by choosing a sequence (k_n)_{n\ge 1} of continuous functions converging to k in the sense of L_2 norm, i.e.
\lim_{n\to\infty}\int_0^1(k_n(x)-k(x))^2\text{d}x=0.
Such a sequence could be constructed using the Fourier decomposition of k into sines for example.
Then for any n\ge 1 define T_nf(x):=\int_0^1k_n(x,y)f(y)dy.
Fix m\ge 1 and for brevity denote S=T_m.
First we show that \text{Im} S\subseteq C[0,1]. Let f\in L_2[0,1]. We aim to show that Sf is a continuous function. Indeed, let x\in [0,1] and (x_n)_{n\ge 1} be a sequence in [0,1] converging to x. Then, if we denote g_n(y)=k_m(x_n,y), we have
\lim_{n\to\infty}g_n(y)=k_m(x,y)
Moreover, if \displaystyle C:=\sup_{x\in[0,1]}|k_m(x,y)| (finite, since k_m is continuous), then |g_n(y)|=|k_m(x_n,y)|\le C. Thus Dominated convergence theorem is applicable, hence
\lim_{n\to\infty}Sf(x_n)=\lim_{n\to\infty}\int_0^1g_n(y)f(y)\text{d}y=\int_0^1k_m(x,y)f(y)\text{d}y=Sf(x).
So S:L_2[0,1]\to C[0,1].
Next, equip C[0,1] with the \sup norm. We will prove S is compact. To this end, we take a sequence (f_n)_{n\ge 1}\subset L_2[0,1], such that \|f_n\|_2\le 1 for all n, and we are going to prove that (Sf_n)_{n\ge 1} has a convergent subsequence in C[0,1]. The criterion telling us whether this is true is the Arzela-Ascoli theorem.
From the Cauchy-Bunyakovsky-Schwartz inequality we obtain |Sf_n(x)|=\left|\int_0^1 k_m(x,y)f_n(y)\text{d}y\right|\le \sqrt{\int_0^1k_m(x,y)^2\text{d}y\int_0^1f_n(y)^2\text{d}y}\le C
(for \displaystyle C:=\sup_{x\in[0,1]}|k_m(x,y)|) and thus the sequence (Sf_n)_{n\ge 1} is uniformly bounded.
Moreover, let \varepsilon>0. Then there is some \delta>0 such that whenever |x_1-x_2|< \delta, then |k_m(x_1,y)-k_m(x_2,y)|<\varepsilon for all y\in[0,1] (uniform continuity of continuous functions).Thus for any n and any x_1,x_2, such that |x_1-x_2|<\delta we have
|Sf_n(x_1)-Sf_n(x_2)|=\left|\int_0^1(k_m(x_1,y)-k_m(x_2,y))f_n(y)\text{d}y\right|
\le \sqrt{\int_0^1(k_m(x_1,y)-k_m(x_2,y))^2\text{d}y \int_0^1 f_n(y)^2\text{d}y}\le \varepsilon
This shows that the sequence (Sf_n)_{n\ge 1} is equicontinuous. Thus it contains convergent subsequence.
This concludes the proof that (T_m)_{m\ge 1} is a sequence of compact operators (acting from L_2[0,1] to C[0,1]). Moreover, for any f\in C[0,1], \|f\|_{\infty}\ge \|f\|_2, which means that the L_2 topology on C[0,1] is weaker than the uniform topology. Hence the operators T_m, m\ge 1, are compact, even if we consider them from L_2[0,1] to L_2[0,1].
It remains to observe that T_m converges to T in the operator norm.
Let f\in L_2[0,1]. Then
\|T_mf-Tf\|_2^2=\int_0^1\left(\int_0^1(k(x,y)-k_m(x,y))f(y)\text{d}y\right)^2\text{d}x\le
\int_0^1\int_0^1(k(x,y)-k_m(x,y))^2\text{d}y\int_0^1f(y)^2\text{d}y\text{d}x=\int_0^1\int_0^1(k(x,y)-k_m(x,y))^2\text{d}y\text{d}x\int_0^1f(y)^2\text{d}y=
\|k_m-k\|_2^2\|f\|_2^2.
This shows that
\|T_m-T\|\le \|k_m-k\|_2,
and since (k_m)_{m\ge 1} was taken to converge to k in L_2, then T_m converges to T.
The proof is finished, by saying, that the space of compact operators is closed, and as T_m are compact, so is T.
Formally, let k\in L_2([0,1]\times[0,1]) and T:L_2[0,1]\to L_2[0,1] be defined by
Tf(x):=\int_0^1k(x,y)f(y)dy.
Then T is compact operator.
Although the proof may seem a bit long at first sight, it is proceeds naturally, utilizing approximation of k with continuous functions.
\textbf{Proof}. We approximate k with a sequence of continuous functions, then prove that each of the corresponding operators has image contained in C[0,1], then show that each of these operators is compact when C[0,1] is equipped with the \sup norm. This will be done via the Arzela-Ascoli theorem. Finally show that these converge uniformly (in the operator norm) to the original operator, and since they are compact, the original must be as well.
The Cauchy-Bunyakovsky-Schwartz inequality is used throughout the proof.
Thus we start by choosing a sequence (k_n)_{n\ge 1} of continuous functions converging to k in the sense of L_2 norm, i.e.
\lim_{n\to\infty}\int_0^1(k_n(x)-k(x))^2\text{d}x=0.
Such a sequence could be constructed using the Fourier decomposition of k into sines for example.
Then for any n\ge 1 define T_nf(x):=\int_0^1k_n(x,y)f(y)dy.
Fix m\ge 1 and for brevity denote S=T_m.
First we show that \text{Im} S\subseteq C[0,1]. Let f\in L_2[0,1]. We aim to show that Sf is a continuous function. Indeed, let x\in [0,1] and (x_n)_{n\ge 1} be a sequence in [0,1] converging to x. Then, if we denote g_n(y)=k_m(x_n,y), we have
\lim_{n\to\infty}g_n(y)=k_m(x,y)
Moreover, if \displaystyle C:=\sup_{x\in[0,1]}|k_m(x,y)| (finite, since k_m is continuous), then |g_n(y)|=|k_m(x_n,y)|\le C. Thus Dominated convergence theorem is applicable, hence
\lim_{n\to\infty}Sf(x_n)=\lim_{n\to\infty}\int_0^1g_n(y)f(y)\text{d}y=\int_0^1k_m(x,y)f(y)\text{d}y=Sf(x).
So S:L_2[0,1]\to C[0,1].
Next, equip C[0,1] with the \sup norm. We will prove S is compact. To this end, we take a sequence (f_n)_{n\ge 1}\subset L_2[0,1], such that \|f_n\|_2\le 1 for all n, and we are going to prove that (Sf_n)_{n\ge 1} has a convergent subsequence in C[0,1]. The criterion telling us whether this is true is the Arzela-Ascoli theorem.
From the Cauchy-Bunyakovsky-Schwartz inequality we obtain |Sf_n(x)|=\left|\int_0^1 k_m(x,y)f_n(y)\text{d}y\right|\le \sqrt{\int_0^1k_m(x,y)^2\text{d}y\int_0^1f_n(y)^2\text{d}y}\le C
(for \displaystyle C:=\sup_{x\in[0,1]}|k_m(x,y)|) and thus the sequence (Sf_n)_{n\ge 1} is uniformly bounded.
Moreover, let \varepsilon>0. Then there is some \delta>0 such that whenever |x_1-x_2|< \delta, then |k_m(x_1,y)-k_m(x_2,y)|<\varepsilon for all y\in[0,1] (uniform continuity of continuous functions).Thus for any n and any x_1,x_2, such that |x_1-x_2|<\delta we have
|Sf_n(x_1)-Sf_n(x_2)|=\left|\int_0^1(k_m(x_1,y)-k_m(x_2,y))f_n(y)\text{d}y\right|
\le \sqrt{\int_0^1(k_m(x_1,y)-k_m(x_2,y))^2\text{d}y \int_0^1 f_n(y)^2\text{d}y}\le \varepsilon
This shows that the sequence (Sf_n)_{n\ge 1} is equicontinuous. Thus it contains convergent subsequence.
This concludes the proof that (T_m)_{m\ge 1} is a sequence of compact operators (acting from L_2[0,1] to C[0,1]). Moreover, for any f\in C[0,1], \|f\|_{\infty}\ge \|f\|_2, which means that the L_2 topology on C[0,1] is weaker than the uniform topology. Hence the operators T_m, m\ge 1, are compact, even if we consider them from L_2[0,1] to L_2[0,1].
It remains to observe that T_m converges to T in the operator norm.
Let f\in L_2[0,1]. Then
\|T_mf-Tf\|_2^2=\int_0^1\left(\int_0^1(k(x,y)-k_m(x,y))f(y)\text{d}y\right)^2\text{d}x\le
\int_0^1\int_0^1(k(x,y)-k_m(x,y))^2\text{d}y\int_0^1f(y)^2\text{d}y\text{d}x=\int_0^1\int_0^1(k(x,y)-k_m(x,y))^2\text{d}y\text{d}x\int_0^1f(y)^2\text{d}y=
\|k_m-k\|_2^2\|f\|_2^2.
This shows that
\|T_m-T\|\le \|k_m-k\|_2,
and since (k_m)_{m\ge 1} was taken to converge to k in L_2, then T_m converges to T.
The proof is finished, by saying, that the space of compact operators is closed, and as T_m are compact, so is T.
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