Ch4. Hilbert Space: An Introduction (Ex.21 ~ Ex.28)

(a)-1 Show, by an example, that weak convergence does not imply strong convergence.

proof ① \((R^nf,g)=(f,L^ng)\) in \(\ell^2(\mathbb{N})\)

Let \(H=\ell^2(\mathbb{N})\), \(T\) be a zero operator, and \(T_n=R^n\) where \(R\) is the right-shift operator : \(T_n(f)=(0,\cdots,0,a_1,a_2,\cdots)\) where \(f=(a_1,a_2,\cdots)\in \ell^2(\mathbb{N})\). Let \(L\) be the left-shift operator and \(g=(b_1,b_2,\cdots)\), then \[ (T_nf,g)=\sum_{k=1}^\infty \overline{a_k}b_{n+k}=(f,L^ng). \]

② \(\|L^ng\|_H\to 0\), \((T_nf,g)\to 0\)

Due to Cauchy-Schwartz inequality, \(|(T_nf,g)|\le \|f\|_H\|L^ng\|_H\).
\(f,g\in \ell^2(\mathbb{N})\), so \(\|f\|<\infty\) and \(\|g\|<\infty\).
\[ \|L^ng\|_H^2=\sum_{k=1}^\infty |b_k|^2-\sum_{k=1}^n|b_k|^2, \] so \(\lim_{n\to\infty}\|L^ng\|_H=0\) and thus \(\lim_{n\to\infty}(T_nf,g)=0\).

③ \(\|T_nf\|_H\not\to 0\)

However, \(\|T_nf\|_H=\|f\|_H\) for all \(f\in \ell^2(\mathbb{N})\). So \(\lim_{n\to\infty}\|T_nf-0\|\ne 0\). \(T_nf-Tf\)

(a)-2 Show that strong convergence does not imply convergence in the norm.

proof Let \(T_n=L^n\). We’ve shown that \(\lim_{n\to\infty}\|L^nf\|_H=0\) for all \(f\in \ell^2(\mathbb{N})\) in (a)-1. So, \(\|T_nf-0f\|_H\to 0\).

Note that \(\|T_n\|=\sup_{\|u\|_H=1}\|T_nu\|_H\). When \(e_k=(\cdots,1,0,0,\cdots)\in H\), \(\|T_n\|\ge \|T_ne_{n+1}\|_H=1\) so \(\|T_n\|\ge 1\).

Then, \(\lim_{n\to\infty}\|T_n-0\|\) can’t be \(0\). So, \(\lim_{n\to\infty}\|T_n-0\|\ne 0\).

(b) Show that for any bounded operator \(T\), there is a sequence \(\{T_n\}\) of bounded operators of finite rank such that \(T_n\to T\) strongly

proof ① def. of \(P_n\), \(T_n\)

Let \(\{e_k\}_{k=1}^\infty\) be an orthonormal basis for \(H\) and \(\alpha_k\in\mathbb{C}\). Let \(P_n\) be the orthogonal projection onto the first ‘\(n\)’ coordinates : \(P_n\left(\sum_{k=1}^\infty \alpha_ke_k\right)=\sum_{k=1}^n \alpha_ke_k\). Let \(T_n=P_nT\). For \(f\in H\), let \(Tf=\sum_{k=1}^\infty \alpha_ke_k\in H\).

② \(T_n\) is bounded

\[ \|T_nf\|_H=\|P_n(Tf)\|_H\le \|Tf\|_H\ \ (\because P_n\ \text{is an orthogonal projection}) \le \|T\|\,\|f\|_H. \] \(T\) is bounded, so \(\|T\|<\infty\) and thus \(T_n\) is bounded

③ \(\|T_nf-Tf\|_H\to 0\)

Then, \[ \|T_nf-Tf\|_H=\left\|\sum_{k=1}^n \alpha_ke_k-\sum_{k=1}^\infty \alpha_ke_k\right\|_H\to 0\ \ \text{as}\ n\to\infty\ \ (\text{Thm 4.2.3 iii}). \] This holds for all \(f\in H\), so \(T_n\to T\) strongly.

(a)-1 Show that if \(T\) is an isometry, then \((Tf,Tg)=(f,g)\) for all \(f,g\in H\).

proof
\[ \begin{aligned} (Tf,Tg)&=\frac14\left\{\|Tf+Tg\|_{H}^2-\|Tf-Tg\|_{H}^2+i\|iTf+Tg\|_{H}^2-i\|iTf-Tg\|_{H}^2\right\}\\ &=\frac14\left\{\|T(f+g)\|_{H}^2-\|T(f-g)\|_{H}^2+i\|T(if+g)\|_{H}^2-i\|T(if-g)\|_{H}^2\right\}\\ &=\frac14\left\{\|f+g\|_{H}^2-\|f-g\|_{H}^2+i\|if+g\|_{H}^2-i\|if-g\|_{H}^2\right\}\\ &=\frac14\Big\{(\|f\|_{H}^2+(f,g)+(g,f)+\|g\|_{H}^2)-(\|f\|_{H}^2-(f,g)-(g,f)+\|g\|_{H}^2)\\ &\qquad\quad+i(\|if\|_{H}^2+(if,g)+(g,if)+\|g\|_{H}^2)-i(\|if\|_{H}^2-(if,g)-(g,if)+\|g\|_{H}^2)\Big\}\\ &=\frac14\left\{2(f,g)+2(g,f)+i\left(\frac1i(f,g)+\frac1{-i}(g,f)\right)+i\left(\frac1i(f,g)+\frac1{-i}(g,f)\right)\right\}\\ &=\frac14\left\{2(f,g)+2(g,f)+2(f,g)-2(g,f)\right\}=(f,g). \end{aligned} \]

(a)-2 Show that \(T^*T=I\)

proof
\((Tf,Tg)=(f,T^*Tg)=(f,Ig)\). So, \(T^*T=I\).

(b) \(T\) is isometry and surjective. Show that \(T\) is unitary & \(TT^*=I\)

proof
When \(f\ne g\in H\), \(\|T(f)-T(g)\|_{H}=\|T(f-g)\|_{H}=\|f-g\|_{H}\ne0.\) Since \(T(f)\ne T(g)\), \(T\) is injective (1-1).
\(T\) is surjective, then \(T\) is bijective \(\Rightarrow T^{-1}\) exists.
\(T\) is linear, bijective, and isometry, so \(T\) is unitary.
\(T^*T=I\), so \(T^*=T^{-1}\) and thus \(TT^*=TT^{-1}=I\).

(c) Give an example of an isometry that is not unitary

proof
Let \(H=\ell^2(\mathbb{N})\) and \(T\) be the right-shift operator: for \(a\in\ell^2(\mathbb{N})\) with \(a=(a_1,a_2,\cdots)\), \(T(a)=(0,a_1,a_2,\cdots)\).
Since \(\|Ta\|_{H}^2=\sum_{k=1}^\infty|a_k|^2=\|a\|_{H}^2\), \(\|Ta\|_{H}=\|a\|_{H}\). So \(T\) is isometry.
When \(b=(b_1,b_2,\cdots)\in\ell^2(\mathbb{N})\) such that \(b_1\ne0\), there is no such \(a'\in\ell^2(\mathbb{N})\) that \(b=T(a)\). So, \(T\) is not surjective (onto).

(d) Show that if \(T^*T\) is unitary, then \(T\) is isometry.

proof
① \(\|Tf\|\le\|f\|\)
\[ \begin{aligned} \|Tf\|^2&=(Tf,Tf)=(f,T^*Tf)\le\|f\|\ \|T^*Tf\|=\|f\|^2,\ \text{so }\|Tf\|\le\|f\|. \end{aligned} \]

② \(\|f\|\le\|Tf\|\)
\[ \begin{aligned} \|f\|^2&=\|T^*Tf\|^2\ \ (\because\ T^*T\ \text{is unitary})\\ &=(T^*Tf,T^*Tf)=(Tf,T(T^*Tf))\\ &\le\|Tf\|\ \|T(T^*Tf)\|\ \ (\because\ \text{Cauchy})\\ &\le\|Tf\|\ \|T^*Tf\|\ \ (\because\ ①)\\ &=\|Tf\|\ \|f\|\ \ (\because\ T^*T\ \text{is unitary}), \end{aligned} \] so \(\|f\|\le\|Tf\|\).

③ \(\therefore\ \|Tf\|=\|f\|\) and thus \(T\) is isometry.

cf) due to (a) and (b), unitary \(U:H\to H'\) implies \(U^*U=I.\) When \(T:H\to H\) is unitary, \(TT^*=T^*T=I\).

(1) Let \(S_N=\sum_{k=-N}^{N}T_k\). Show that \(S_N(x)\) converges (in norm) as \(N\to\infty\).

proof
① \(P_k\), \((P_ky,P_{k'}z)=0\).

Let \(P_k\) be an orthogonal projection onto the range of \(T_k\): \(y=P_ky+(I-P_k)y\) where \(y\in\mathcal{H}\) and \(P_ky\in\{T_k f:\,f\in\mathcal{H}\}\). Note that \(T_kx\perp (I-P_k)y\).

For \(y\in\mathcal{H}\), there is \(\tilde y\in\mathcal{H}\) s.t. \(P_ky=T_k\tilde y\). Then, \[(P_ky,P_{k'}z)=(T_k\tilde y,T_{k'}\tilde z)=(\tilde y,T_k^*T_{k'}\tilde z)=0.\] So \((P_ky,P_{k'}z)=0\) for \(k\ne k'\) and all \(y,z\in\mathcal{H}\).

② \(Q_k\), \((Q_kx,Q_{k'}z)=0\).

Let \(Q_k\) be an orthogonal projection onto the range of \(T_k^*\): \(x=Q_kx+(I-Q_k)x\) where \(x\in\mathcal{H}\) and \(Q_kx\in\{T_k^*y:\,y\in\mathcal{H}\}\). Then \(T_k^*y\perp (I-Q_k)x\).

For \(x\in\mathcal{H}\), there is \(\tilde x\in\mathcal{H}\) s.t. \(Q_kx=T_k^*\tilde x\). \[(Q_kx,Q_{k'}z)=(T_k^*\tilde x,T_{k'}^*\tilde z)=(\tilde x,T_kT_{k'}^*\tilde z)=0.\] So \((Q_kx,Q_{k'}z)=0\) for \(k\ne k'\) and \(\forall\,z_1,z_2\in\mathcal{H}\).

③ \(T_kx\perp T_{k'}y\).

When \(k\ne k'\) and \(x,y\in\mathcal{H}\), \[(T_kx,T_{k'}y)=(x,T_k^*T_{k'}y)=0.\] So \(T_kx\perp T_{k'}y\) for all \(x,y\in\mathcal{H}\).

④ \((T_kx,y)=(T_kQ_kx,P_ky)\).

\[\begin{aligned} (T_kx,y) &=(T_kx,P_ky+(I-P_k)y) \\ &=(T_kx,P_ky)\qquad(\because\,T_kx\perp(I-P_k)y) \\ &=(x,T_k^*P_ky) \\ &=(Q_kx+(I-Q_k)x,T_k^*P_ky) \\ &=(Q_kx,T_k^*P_ky)\qquad(\because\,(I-Q_k)x\perp T_k^*P_ky) \\ &=(T_kQ_kx,P_ky). \end{aligned}\]

⑤-1 \(\left|\sum_{k\in F}(T_kx,y)\right|\le\left(\sum_{k\in F}\|Q_kx\|^2\right)^{1/2}\left(\sum_{k\in F}\|P_ky\|^2\right)^{1/2}\).

Let \(F\) be any finite subset of \(\mathbb{N}\). Then, \[\begin{aligned} \left|\sum_{k\in F}(T_kx,y)\right| &\le \sum_{k\in F}|(T_kx,y)| \\ &= \sum_{k\in F}|(T_kQ_kx,P_ky)|\qquad(\because\,④) \\ &\le \sum_{k\in F}\|T_kQ_kx\|\,\|P_ky\|\qquad(\because\,\text{Cauchy–Schwarz inequality in }\mathcal{H}) \\ &\le \sum_{k\in F}\|Q_kx\|\,\|P_ky\|\qquad(\because\,\|T_k\|\le 1) \\ &\le \left(\sum_{k\in F}\|Q_kx\|^2\right)^{1/2}\left(\sum_{k\in F}\|P_ky\|^2\right)^{1/2}\qquad(\because\,\text{Cauchy–Schwarz inequality in }\mathbb{R}). \end{aligned}\]

⑤-2 \(p_k=\|P_ky\|^{-1}P_ky,\ \ q_k=\|Q_kx\|^{-1}Q_kx\) orthonormal set.

Let \(p_k=\|P_ky\|^{-1}P_ky\). \[(p_k,p_{k'})=\|P_ky\|^{-1}\|P_{k'}y\|^{-1}(P_ky,P_{k'}y)=0,\ \ \text{and }\|p_k\|=1.\] So \(\{p_k\}_k\) is an orthonormal “set” in \(\mathcal{H}\).

Let \(q_k=\|Q_kx\|^{-1}Q_kx\). \[(q_k,q_{k'})=\|Q_kx\|^{-1}\|Q_{k'}x\|^{-1}(Q_kx,Q_{k'}x)=0,\ \ \text{and }\|q_k\|=1.\] So \(\{q_k\}_k\) is an orthonormal set in \(\mathcal{H}\).

⑤-3 \((y,p_k)=\|P_ky\|,\ \ (x,q_k)=\|Q_kx\|\).

\[\begin{aligned} (y,p_k) &=\|P_ky\|^{-1}(y,P_ky) =\|P_ky\|^{-1}(P_ky+(I-P_k)y,P_ky) \\ &=\|P_ky\|^{-1}(P_ky,P_ky) =\|P_ky\|, \end{aligned}\]

\[\begin{aligned} (x,q_k) &=\|Q_kx\|^{-1}(x,Q_kx) =\|Q_kx\|^{-1}(Q_kx+(I-Q_k)x,Q_kx) \\ &=\|Q_kx\|^{-1}(Q_kx,Q_kx) =\|Q_kx\|. \end{aligned}\]

⑤-4 \(\left|\sum_{k\in F}(T_kx,y)\right|\le \|x\|\,\|y\|\).

Since \(\sum_{k\in F}(y,p_k)^2\le\sum_{k\in\mathbb{N}}(y,p_k)^2\le\|y\|^2\) (Bessel’s inequality p.166), \[\sum_{k\in F}\|P_ky\|^2=\sum_{k\in F}(y,p_k)^2\le\|y\|^2,\quad\text{and similarly }\sum_{k\in F}\|Q_kx\|^2\le\|x\|^2.\] So, \[\left|\sum_{k\in F}(T_kx,y)\right|\le\left(\sum_{k\in F}\|Q_kx\|^2\right)^{1/2}\left(\sum_{k\in F}\|P_ky\|^2\right)^{1/2}\le \|x\|\,\|y\|.\]

⑥ \(\left\|\sum_{k\in F}T_kx\right\|\le \|x\|\).

Let \(w=\sum_{k\in F}T_kx\). When \(w=0\), \(\|w\|=0\le\|x\|\) (trivial).

When \(w\ne 0\), let \(y=\dfrac{w}{\|w\|}\). \((w,y)=\left(w,\frac{w}{\|w\|}\right)=\|w\|.\) Also, \(|(w,y)|=\left|\left(\sum_{k\in F}T_kx,y\right)\right|\le \|x\|\,\|y\|\ (\because\,⑤)=\|x\|.\) Thus, \(\|w\|\le\|x\|\).

⑦ \(\sum_{k\in F}\|T_kx\|^2=\left\|\sum_{k\in F}T_kx\right\|^2\), so \(\sum_{k\in F}\|T_kx\|^2\le\|x\|^2\).

Let \(k_1,k_2\in F\) s.t. \(k_1\ne k_2\). \[\begin{aligned} \|T_{k_1}x+T_{k_2}x\|^2 &=(T_{k_1}x+T_{k_2}x,\ T_{k_1}x+T_{k_2}x) \\ &=\|T_{k_1}x\|^2+(T_{k_1}x,T_{k_2}x)+(T_{k_2}x,T_{k_1}x)+\|T_{k_2}x\|^2 \\ &=\|T_{k_1}x\|^2+\|T_{k_2}x\|^2\qquad(\because\,T_{k_1}^*T_{k_2}=0). \end{aligned}\]

Repeating the process, \(\sum_{k\in F}\|T_kx\|^2=\left\|\sum_{k\in F}T_kx\right\|^2\). Since \(\left\|\sum_{k\in F}T_kx\right\|\le\|x\|\ (\because\,⑥)\), \(\sum_{k\in F}\|T_kx\|^2\le\|x\|^2\).

⑧ \(u_n(x)=\sum_{k=-n}^{n}\|T_kx\|^2\) converges in \(\mathbb{R}\)

Let \(u_n(x)=\sum_{k=-n}^{n}\|T_kx\|^2\) for \(x\in\mathcal{H}\). Since \(u_n(x)\le\|x\|^2<\infty\), \(\{u_n(x)\}_n\) converges and thus \(\{u_n(x)\}_n\) in \(\mathbb{R}\) is a Cauchy sequence. So, for given \(\varepsilon>0\), there is \(N_\varepsilon\in\mathbb{N}\) such that \(m>n\ge N_\varepsilon\) implies \(|u_m(x)-u_n(x)|=\sum_{n<|k|\le m}\|T_kx\|^2<\varepsilon\).

⑨ \(S_n(x)=\sum_{k=-n}^{n}T_kx\) is Cauchy in \(\mathcal{H}\), \(\therefore\ S_n(x)\to F\)

Using ⑦, \(\sum_{n<|k|\le m}\|T_kx\|^2=\left\|\sum_{n<|k|\le m}T_kx\right\|^2=\left\|\sum_{k=-m}^{m}T_kx-\sum_{k=-n}^{n}T_kx\right\|^2<\varepsilon\) for \(m>n\ge N_\varepsilon\). This holds for all \(\varepsilon>0\), so \(\{S_n(x)\}_n=\left\{\sum_{k=-n}^{n}T_kx\right\}_n\) is a Cauchy sequence in \(\mathcal{H}\).

\(\mathcal{H}\) is complete, so \(\{S_n(x)\}_n\) converges to some \(F_x\in\mathcal{H}\): \(\lim_{n\to\infty}\left\|\sum_{k=-n}^{n}T_kx-F_x\right\|=0\). Note that this holds for all \(x\in\mathcal{H}\).

(2) Show that \(\|T\|\le 1\)

proof
Define \(T:\mathcal{H}\to\mathcal{H}\) as follows: \(T(x)=F_x\).

\(\lim_{n\to\infty}\|S_n(x)-T(x)\|=0\), so for given \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(n\ge N_\varepsilon\) implies \(\|S_n(x)-T(x)\|<\varepsilon\).

Since \(\|S_n(x)\|\le\|x\|\) (⑥), \[\|Tx\|\le\|Tx-S_nx\|+\|S_nx\|\le \varepsilon+\|x\|.\]

\(\|Tx\|\le \varepsilon+\|x\|\) holds for all \(\varepsilon>0\), so \(\|Tx\|\le\|x\|\) and thus \(\|T\|\le 1\).

(1) Show that \(T\) is bounded on \(L^2(\mathbb{R}^d)\) : \(\|T\|\le A\)

proof
① \(\left\{\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|\,dy\right\}^2\le A\,\omega(x)\int_{\mathbb{R}^d}|K(x,y)|\,\omega(y)^{-1}\,|f(y)|^2\,dy\ \text{a.e }x\)

\[\begin{aligned} \int_{\mathbb{R}^d}|K(x,y)f(y)|\,dy &=\int_{\mathbb{R}^d}\Big(\sqrt{|K(x,y)|\omega(y)}\Big)\Big(\sqrt{|K(x,y)|\omega(y)^{-1}|f(y)|^2}\Big)\,dy, \end{aligned}\] so \[\begin{aligned} \left\{\int_{\mathbb{R}^d}|K(x,y)f(y)|\,dy\right\}^2 &=\left\{\int_{\mathbb{R}^d}\Big(\sqrt{|K(x,y)|\omega(y)}\Big)\Big(\sqrt{|K(x,y)|\omega(y)^{-1}|f(y)|^2}\Big)\,dy\right\}^2 \\ &\le \left(\int_{\mathbb{R}^d}|K(x,y)|\omega(y)\,dy\right)\left(\int_{\mathbb{R}^d}|K(x,y)|\omega(y)^{-1}|f(y)|^2\,dy\right) \\ &\Big(\because \text{Cauchy–Schwarz inequality} \Big)\\ &\le A\cdot\omega(x)\int_{\mathbb{R}^d}|K(x,y)|\omega(y)^{-1}|f(y)|^2\,dy\ \text{a.e }x. \end{aligned}\] (\(\because\) Cauchy–Schwarz inequality)

② \(\|Tf\|_2\le A\)

\[\begin{aligned} \|Tf\|_2^2 &=\int_{\mathbb{R}^d}\left|\int_{\mathbb{R}^d}K(x,y)f(y)\,dy\right|^2dx \\ &\le \int_{\mathbb{R}^d}\left\{\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|\,dy\right\}^2dx \\ &\le \int_{\mathbb{R}^d}A\omega(x)\int_{\mathbb{R}^d}|K(x,y)|\omega(y)^{-1}|f(y)|^2\,dy\cdot dx\qquad(\because\,①) \\ &=A\int_{\mathbb{R}^d}\int_{\mathbb{R}^d}\omega(x)|K(x,y)|\omega(y)^{-1}|f(y)|^2\,dx\,dy\qquad(\because\,\text{Tonelli's thm}) \\ &=A\int_{\mathbb{R}^d}\omega(y)^{-1}|f(y)|^2\int_{\mathbb{R}^d}|K(x,y)|\omega(x)\,dx\cdot dy \\ &\le A\int_{\mathbb{R}^d}\omega(y)^{-1}|f(y)|^2\cdot A\omega(y)\,dy\qquad(\because\,(ii)) \\ &=A^2\int_{\mathbb{R}^d}|f(y)|^2\,dy =A^2\|f\|_2^2. \end{aligned}\]

Thus, \(\|Tf\|_2\le A\|f\|_2\) and \(\|T\|\le A\).

(2) \(\int_{\mathbb{R}^d}|K(x,y)|\,dy\le A\ \forall x\) and \(\int_{\mathbb{R}^d}|K(x,y)|\,dx\le A\ \forall y\). Show that \(\|T\|\le A\).

proof
\[\begin{aligned} \left\{\int_{\mathbb{R}^d}|K(x,y)f(y)|\,dy\right\}^2 &=\left\{\int_{\mathbb{R}^d}\sqrt{|K(x,y)|}\,\sqrt{|K(x,y)|}\,|f(y)|\,dy\right\}^2 \\ &\le \left(\int_{\mathbb{R}^d}|K(x,y)|\,dy\right)\left(\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|^2\,dy\right) \\ &\le A\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|^2\,dy.\qquad\sim\text{①} \end{aligned}\]

Then, \[\begin{aligned} \|Tf\|_2^2 &=\int_{\mathbb{R}^d}\left|\int_{\mathbb{R}^d}K(x,y)f(y)\,dy\right|^2dx \\ &\le \int_{\mathbb{R}^d}\left\{\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|\,dy\right\}^2dx \\ &\le \int_{\mathbb{R}^d}A\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|^2\,dy\cdot dx\qquad(\because\,①) \\ &=A\int_{\mathbb{R}^d}\int_{\mathbb{R}^d}|K(x,y)|\,|f(y)|^2\,dx\,dy\qquad(\because\,\text{Tonelli's thm}) \\ &=A\int_{\mathbb{R}^d}|f(y)|^2\int_{\mathbb{R}^d}|K(x,y)|\,dx\cdot dy \\ &\le A\int_{\mathbb{R}^d}|f(y)|^2\cdot A\,dy =A^2\|f\|_2^2. \end{aligned}\]

So, \(\|Tf\|_2\le A\|f\|_2\), and thus \(\|T\|\le A\).

(a) Show that \(T\) is a bounded operator on \(\mathcal{H}\). \((B:\ |x|\le 1\ \text{wlog})\)

proof
① \(\int_B|K(x,y)|dy\le C_1,\ \int_B|K(x,y)|dx\le C_2\)

\[\int_B|K(x,y)|\,dy\le A\int_B|y-x|^{-d+\alpha}dy=A\int_{B-x}|y'|^{-d+\alpha}dy'\le A\int_{2B}|y'|^{-d+\alpha}dy'\quad(2B:\ |z|\le 2).\]

Since \((d-\alpha)<d\), \(|y'|^{-d+\alpha}\) is integrable on \(2B\). Let \(A\int_{2B}|y'|^{-d+\alpha}dy'=C_1\).

Similarly, \(\int_B|K(x,y)|dx\le A\int_{2B}|x'|^{-d+\alpha}dx'=:C_2\) for \(x'=x-y\).

② \(\|Tf\|_{\mathcal{H}}\le \sqrt{C_1C_2}\|f\|_{\mathcal{H}}\), conclusion

\[\begin{aligned} \|Tf\|_{\mathcal{H}}^2 &=\int_B\left|\int_BK(x,y)f(y)\,dy\right|^2dx \\ &\le \int_B\left\{\int_B|K(x,y)||f(y)|\,dy\right\}^2dx\qquad(\because\ \Delta\ \text{ineq}) \\ &=\int_B\left\{\int_B\sqrt{|K(x,y)|}\,\sqrt{|K(x,y)|}\,|f(y)|\,dy\right\}^2dx \\ &\le \int_B\left\{\int_B|K(x,y)|\,dy\right\}\left\{\int_B|K(x,y)||f(y)|^2dy\right\}dx\qquad(\because\ \text{Cauchy}) \\ &\le \int_B C_1\int_B|K(x,y)||f(y)|^2dy\ dx \\ &=C_1\int_B|f(y)|^2\int_B|K(x,y)|\,dx\ dy\qquad(\because\ \text{Tonelli's thm}) \\ &\le C_1\int_B|f(y)|^2dy\cdot C_2 =C_1C_2\|f\|_{\mathcal{H}}^2. \end{aligned}\]

So, \(\|Tf\|_{\mathcal{H}}\le \sqrt{C_1C_2}\|f\|_{\mathcal{H}}\) and \(\|T\|\le \sqrt{C_1C_2}\Rightarrow\) bounded.

(b) Show that \(T\) is compact

proof
Let \[ K_n(x,y)= \begin{cases} K(x,y), & |x-y|\ge 1/n,\\ 0, & \text{o.w}, \end{cases} \qquad\text{and}\qquad T_nf(x)=\int_BK_n(x,y)f(y)\,dy\ (n\in\mathbb{N},\ f\in\mathcal{H}). \]

① \(T_n\) is a Hilbert-Schmidt operator

\[|K_n(x,y)|\le |K(x,y)|\ (\text{when }|x-y|^{-1}\le n)\le A|x-y|^{-d+\alpha}\le A\cdot n^{d-\alpha}.\]

Then, \[\int_B\int_B|K_n(x,y)|^2dxdy\le A^2n^{2d-2\alpha}\int_B\int_Bdxdy,\] so \(\le A^2n^{2d-2\alpha}\cdot C<\infty\).

So, \(K_n(x,y)\in L^2(B\times B)\) and \(T_n\) is a Hilbert-Schmidt operator.

Note that \(T_n\) is consequently compact.

②-1 \(\int_B|K(x,y)-K_n(x,y)|\,dy\le D_n\) with \(\lim_{n\to\infty}D_n=0\)

\[\int_B|K(x,y)-K_n(x,y)|\,dy=\int_{|x-y|<1/n}|K(x,y)|\,dy\le \int_{|x-y|<1/n}A|x-y|^{-d+\alpha}dy=A\int_{|z|<1/n}|z|^{-d+\alpha}dz.\]

\(|z|^{-d+\alpha}\) is integrable on \(\{z:\ |z|<1\}\), so for given \(\varepsilon>0\) there is \(\delta_\varepsilon>0\) such that \(m(E)<\delta_\varepsilon\) implies \(0\le A\int_E|z|^{-d+\alpha}dz<\varepsilon\). (prop 2.1.12 (ii))

Let \(N_\varepsilon\) be the minimum of \(n\in\mathbb{N}\) such that \(m(\{|z|<1/n\})<\delta_\varepsilon\).

Then \(n\ge N_\varepsilon\) implies \(0\le A\int_{|z|<1/n}|z|^{-d+\alpha}dz<\varepsilon\).

This holds for all \(\varepsilon>0\), so \(\lim_{n\to\infty}A\int_{|z|<1/n}|z|^{-d+\alpha}dz=0\).

Denote \(A\int_{|z|<1/n}|z|^{-d+\alpha}dz\) as \(D_n\); \(\lim_{n\to\infty}D_n=0\). Then \(\int_B|K(x,y)-K_n(x,y)|\,dy\le D_n\).

②-2 \(\int_B|K(x,y)-K_n(x,y)|\,dx\le D_n\)

\[\begin{aligned} \int_B|K(x,y)-K_n(x,y)|\,dx&=\int_{|x-y|<1/n}|K(x,y)|\,dx \\ &\le \int_{|x-y|<1/n}A|x-y|^{-d+\alpha}dx \\ &=A\int_{|z|<1/n}|z|^{-d+\alpha}dz':=D_n. \qquad(\because\ ②\text{-}1) \end{aligned}\]

③ \(\|T-T_n\|\le D_n\)

\[\begin{aligned} \|(T-T_n)f\|_{\mathcal{H}}^2 &=\int_B\left|\int_BK(x,y)f(y)\,dy-\int_BK_n(x,y)f(y)\,dy\right|^2dx \\ &=\int_B\left|\int_B(K(x,y)-K_n(x,y))f(y)\,dy\right|^2dx \\ &\le \int_B\left\{\int_B|K(x,y)-K_n(x,y)|\,|f(y)|\,dy\right\}^2dx \\ &=\int_B\left\{\int_B\sqrt{|K(x,y)-K_n(x,y)|}\,\sqrt{|K(x,y)-K_n(x,y)|}\,|f(y)|\,dy\right\}^2dx \\ &\le \int_B\left\{\int_B|K(x,y)-K_n(x,y)|\,dy\right\}\left\{\int_B|K(x,y)-K_n(x,y)||f(y)|^2dy\right\}dx\qquad(\because\ \text{Cauchy}) \\ &\le \int_B D_n\int_B|K(x,y)-K_n(x,y)||f(y)|^2dy\ dx\qquad(\because\ ②) \\ &=D_n\int_B\int_B|K(x,y)-K_n(x,y)||f(y)|^2dxdy\qquad(\because\ \text{Fubini's thm}) \\ &=D_n\int_B|f(y)|^2\int_B|K(x,y)-K_n(x,y)|\,dx\ dy \\ &\le D_n^2\|f\|_{\mathcal{H}}^2\qquad(\because\ ②)\ (f\in\mathcal{H},\ \text{so }\|f\|_{\mathcal{H}}<\infty). \end{aligned}\]

So \(\|(T-T_n)f\|_{\mathcal{H}}\le D_n\|f\|_{\mathcal{H}}\) and \(\|T-T_n\|\le D_n\).

④ conclusion

\(\lim_{n\to\infty}D_n=0\), so \(\lim_{n\to\infty}\|T-T_n\|=0\). Thus, \(T\) is compact (Prop 4.6.1(ii)).

(c) Show that \(T\) is Hilbert-Schmidt operator if \(\alpha>\dfrac{d}{2}\)

proof
\(T\) is Hilbert-Schmidt operator

\[\Leftrightarrow\ K(x,y)\in L^2(B\times B)\ \Leftrightarrow\ A|x-y|^{-d+\alpha}\in L^2(B\times B)\]

\[\Leftrightarrow\ A^2\int_B\int_B|x-y|^{-2d+2\alpha}dxdy<\infty\ \Leftrightarrow\ 2d-2\alpha<d\ \Leftrightarrow\ 2\alpha>d.\]