1. Let E be a Banach space, and let T \in \mathcal L(E) with \|T\|< 1.
(i). Prove that (I − T) is bijective and that \|(I −T)^{-1}\|\leq 1/(1−\|T\|).
- Let S_n=\sum_{k=0}^{n-1}T^k. As \|S_n-S_m\|=\|\sum_{k=m}^{n-1}T^k\|\leq\sum_{k=m}^{n-1}\|T\|^k=\frac{\|T\|^{m}(1-\|T\|^{n-m})}{1-\|T\|}\rightarrow 0 when m,n\rightarrow \infty. Therefore, (S_n) is a Cauchy squence and by the completeness of \mathcal L (E), we have that \exists S\in\mathcal L(E),s.t\enspace S=\lim_{n\rightarrow\infty}S_n. Consider that (I-T)S=S-TS=\sum_{k=0}^\infty T^k-\sum_{k=1}^\infty T^k=I, we have S=(I-T)^{-1}, hence (I-T) is bijective and \|(I-T)^{-1}\|=\|S\|\leq\sum_{k=0}^\infty\|T\|^k=\frac{1}{1-\|T\|}.
(ii). Set S_n = I +T +\cdots+T^{n−1}. Prove that \|S_n −(I −T)^{-1}\|\leq\|T\|^n/(1−\|T\|).
- Notice that \|S_n-(I-T)^{-1}\|=\|S-S_n\|=\|\sum_{k=n}^\infty T^k\|\leq\sum_{k=n}^\infty\|T^k\|=\frac{\|T\|^n}{1-\|T\|}.
2. Let E, F be two Banach spaces. Let A \in\mathcal L(E,F) be a bounded linear operator. Recall that N(A), R(A) and G(A) denote the kernel, range and graph, respectively. Write G=G(A), L =E\times\{o\}. Verify that:
(1) N(A)\times\{o\} = G\cap L,
- Consider N(A)\subseteq E, we have N(A)\times\{o\}\subseteq E\times\{o\}=L. And for all x\in N(A), A(x)=0, therefore (x,0)\in G\Rightarrow N(A)\times\{o\}\subseteq G. Hence N(A)\times\{o\}\subseteq G\cap L. Now consider x\in G\cap L. x\in L gives that x=(x_E,0),x_E\in E while x\in G giving that A(x_E)=0, which shows that x_E\in N(A) and x\in N(A)\times\{o\}, hence G\cap L\subseteq N(A)\times\{o\}. So, we have N(A)\times\{o\} = G\cap L.
(2) E \times R(A) =G+L,
- Consider x\in E\times R(A), we have x=(x_1,A(x_2)),x_1,x_2\in E. As x=(x_2,A(x_2))+(x_1-x_2,0), and (x_2,A(x_2))\in G,(x_1-x_2,0)\in L, we know that x\in G+L, hence E\times R(A)\subseteq G+L. Now consider x=(x_1,A(x_1))+(x_2,0)\in G+L, x_1,x_2\in E. Notice that x=(x_1+x_2,A(x_1)) and x_1+x_2\in E,A(x_1)\in R(A), we get G+L\subseteq E\times R(A). Therefore, E \times R(A) =G+L.
(3) \{o\} \times N(A^*) = G^\perp \cap L^\perp,
- Consider f=(0_{E^*},f_{F^*})\in\{o\}\times N(A^*)\subseteq E^*\times F^*\cong(E\times F)^*,f_{F^*}\in N(A^*)\subseteq F^*, we have A^*(f_{F^*})=0. We consider f as a funtional in (E\times F)^*. So \forall y\in G,z\in L,\langle f,y\rangle=\langle(0_{E^*},f_{F^*}),(y_E,A(y_E))\rangle=\langle f_{F^*},A(y_E)\rangle=\langle A^*(f_{F^*}),y_E\rangle=0;\langle f,z\rangle=\langle (0,f_{F^*}),(z_E,0)\rangle=0, which means f\in G^\perp and f\in L^\perp. Therefore \{o\} \times N(A^*) \subseteq G^\perp \cap L^\perp. Now consider f=(f_{E^*},f_{F^*})\in G^\perp \cap L^\perp\subseteq (E\times F)^*\cong E^*\times F^*. f\in L^\perp gives that \langle (f_{E^*},f_{F^*}),(z_E,0)\rangle=0,\forall z_E\in E\Rightarrow f_{E^*}=0_{E^*} and f\in G^\perp gives that \langle(0_{E^*},f_{F^*}),(y_E,A(y_E))\rangle=\langle f_{F^*},A(y_E)\rangle=\langle A^*(f_{F^*}),y_E\rangle=0,\forall y_E\in E\Rightarrow f_{F^*}\in N(A^*). So, we have f\in \{o\} \times N(A^*)\Rightarrow G^\perp\cap L^\perp\subseteq\{o\}\times N(A^*). Therefore, we get \{o\} \times N(A^*) = G^\perp \cap L^\perp.
(4) R(A^*) \times F^* = G^\perp+L^\perp.
- Consider f=(A^*(f_1),f_2)\in R(A^*)\times F^*\subseteq E^*\times F^*\cong (E\times F)^*,f_1\in F^*,f_2\in E^*, we have f=(A^*(f_1),-f_1)+(0_{E^*},f_1+f_2). Notice that \forall y\in G,z\in L,\langle(A^*(f_1),-f_1),(y_E,A(y_E))\rangle=\langle f_1,A(y_E)\rangle+\langle-f_1,A(y_E)\rangle=0;\langle (0_{E^*},f_1+f_2),(z_E,0)\rangle=0, we get f\in G^\perp+L^\perp, hence R(A^*) \times F^* \subseteq G^\perp+L^\perp. Now consider f=f^G+f^L\in G^\perp+L^\perp. f^G\in G^\perp gives that \langle(f^G_{E^*},f^G_{F^*}),(y_E,A(y_E))\rangle=\langle f^G_{E^*},y_E\rangle+\langle A^*( f^G_{F^*}),y_E\rangle=0,\forall y_E\in E\Rightarrow f^G_{E^*}=-A^*(f^G_{F^*})=A^*(-f^G_{F^*}), and f^L\in L^\perp gives that \langle(f^L_{E^*},f^L_{F^*}),(z_E,0)\rangle=\langle f^L_{E^*},z_E\rangle=0,\forall z_E\in E\Rightarrow f^L_{E^*}=0_{E^*}. So f=(A^*(-f^G_{F^*}),f^G_{F^*})+(0_{E^*},f^L_{F^*})=(A^*(-f^G_{F^*}),f^G_{F^*}+f^L_{F^*})\in R(A^*)\times F^*. Therefore, R(A^*) \times F^* = G^\perp+L^\perp.
3. Let (E,\|\cdot\|) be a normed vector space. For x, y\in E, define (x, y) = \frac{1}{2}(\|x +y\|^2 −\|x\|^2 −\|y\|^2). Prove that (x,y) defines an inner product if and only if the norm \|\cdot\| satisfies the parallelogram law \|x +y\|^2 +\|x−y\|^2 = 2\|x\|^2 +2\|y\|^2, \forall x,y \in E.
- P\Rightarrow Q: Consider (x,y) to be an inner product, we have (x,y)=-(x,-y), which means \|x+y\|^2-\|x\|^2-\|y\|^2=-\|x-y\|^2+\|x\|^*+\|y\|^2. So \|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.
P\Leftarrow Q: Consider we have \|x +y\|^2 +\|x−y\|^2 = 2\|x\|^2 +2\|y\|^2, we also can get (x,y)=-(-x,y),\forall x,y\in E. As (x,y)=(y,x) and (x,x)=\|x\|^2, it's obvious that (\cdot,\cdot) is symmetric and positive definite. Now we consider the linearity. First, for all x,z\in E, (x,2z)=\frac{1}{2}(\|x+2z\|^2-\|x\|^2-\|2z\|^2)=\frac{1}{2}[(\|x+2z\|^2+\|x\|^2)-2\|x\|^2-\|2z\|^2]=\frac{1}{2}[\frac{1}{2}(\|2x+2z\|^2+\|2z\|^2)-2\|x\|^2-\|2z\|^2]=\|x+z\|^2-\|x\|^2-\|z\|^2=2(x,z). Then, for all x,x',y\in E, (x,y)+(x',y)=2(x,\frac{1}{2}y)+2(x',\frac{1}{2}y)=\|x+\frac{1}{2}y\|^2-\|x\|^2-\|\frac{1}{2}y\|^2+\|x'+\frac{1}{2}y\|^2-\|x'\|^2-\|\frac{1}{2}y\|^2=\frac{1}{2}(\|x+x'+y\|^2+\|x-x'\|^2)-\frac{1}{2}(\|x+x'\|^2+\|x-x'\|^2)-\frac{1}{2}\|y\|^2=\frac{1}{2}(\|x+x'+y\|^2-\|x+x'\|^2-\|y\|^2)=(x+x',y). So, for all k\in\mathbb Q,x,y\in E, we have (kx,y)=k(x,y). Then, by the continuity of \|\cdot\|, we have for all \lambda\in\R, (\lambda x,y)=\lambda(x,y). So, we have proved that (x,y) is an inner product.
Q.E.D.
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