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The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ...

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As $T$ is bounded and self-adjoint with norm $1$, its spectrum $\sigma(T)$ is a compact subset of $[-1,1]$, and its spectral radius $r(T)$ equals $1$. Hence, either $-1 \in \sigma(T)$ or $1 \in \sigma(T)$. If $-1 \in \sigma(T)$, then $0 \in \sigma(I + T)$, and so $I + T$ is not invertible. If $1 \in \sigma(T)$, then $0 \in \sigma(I ... 4 Here is another answer that is suitable for the poster’s background. We make use of the fact that for a self-adjoint bounded operator$ T $on a Hilbert space$ \mathcal{H} $, we have $$\| T \|_{B(\mathcal{H})} = \sup(\{ |\langle T(h),h \rangle_{\mathcal{H}}| \mid h \in \mathbb{S}(\mathcal{H}) \}),$$ where$ \mathbb{S}(\mathcal{H}) $denotes the unit ... 3 Consider $$T:\ell^2\rightarrow\ell^2\\T(x_n;\ n\in\mathbb{N}^+)=\left(\frac{x_n}{n};\ n\in\mathbb{N}^+\right)$$ Notice that$\|T\|\leq1$, but$\left(\frac{1}{n};\ n\in\mathbb{N}^+\right)\notin T(\ell^2)$But$\forall g\in \ell^{2*}\ (g\circ T=0\rightarrow g=0)$because$T(\ell^2)$contains a Hilbert basis. So the answer to your question is no. 2 Let$ \mathcal{H} $be a Hilbert space and$ (\mathbf{e}_{i})_{i \in I} $a Hilbert basis of$ \mathcal{H} $. If$ f $is a linear mapping from$ \mathcal{H} $to$ \mathbb{C} $, then$ f $is continuous if and only if $$(\spadesuit) \qquad (f(\mathbf{e}_{i}))_{i \in I} \in {\ell^{2}}(I).$$ Hence, although a linear functional on$ \mathcal{H} $is uniquely ... 2 As KCd said in a comment, a continuous linear functional on a Hilbert space$H$is uniquely determined by its action on a given orthonormal basis. This follows since, if$M$is an orthonormal basis of$H$, then for any$x\in H$we have $$f(x)=\sum_{m\in M}\langle x,m\rangle f(m)$$ which follows from continuity. If$\dim H=\infty$and we do not require$f$to ... 2 The fact that such a$ * $-homomorphism exists is the statement of the Gelfand-Naimark-Segal (GNS) Construction. 2 $$\langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle\forall \alpha\\ \langle x,x\rangle+2\alpha\langle x,y\rangle+\alpha^2\langle y,y\rangle\geq\langle x,x\rangle\forall\alpha\\ 2\alpha \langle x,y\rangle+\alpha^2\langle y,y\rangle \geq0\forall\alpha$$ But the left-hand side equals zero if$\alpha=0$and if$\alpha=-2\langle x,y\rangle/\langle ...

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Let $\varphi: \mathcal{H} \to \mathbb{C}$ be any (continuous) linear functional on $\mathcal{H}$. Then the composition $$\varphi \circ T = \left\{ \begin{matrix} {\ell^{2}}(\mathbb{N}) & \to & \mathbb{C} \\ (a_{n})_{n \in \mathbb{N}} & \mapsto & \sum_{n \in \mathbb{N}} a_{n} \varphi(f_{n}) ... 1 You need to show that (\mathrm{ran}\, A)^\perp = \{0\} if and only if \ker A = \{0\}. The key is that A^* is injective on the range of A: if A^*Ax = 0 then$$ 0 = \langle x,A^*Ax \rangle = \langle Ax , Ax \rangle = \|Ax\|^2 $$so that Ax = 0. Similarly, A is injective on the range of A^*. This means that$$\ker A = \ker A^*A \quad ...

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The image of $T$ is also closed: Take $T(x_k)\rightarrow y_0 \in H_2$. Then $T(x_k)$ is Cauchy. Since $T$ is bounded below, the sequence $x_k$ is also Cauchy. So $x_k\rightarrow x_0$. So $T(x_0)=y_0$.

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Call $Px$ the orthogonal projection of $x$ onto $M$, then it's easy to see that $\| x-Px\|= \inf_{y\in M}\| x-y\|$. On the other hand we have that $\max_{y\in M^\perp, \| y\|=1} |\langle x,y\rangle|$ is just the norm of the linear functional $l(y)=\langle x,y\rangle$ when viewed as $l\colon M^\perp \to \mathbb{R}$. Since $Px \perp M^\perp$ we have that  ...

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One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book. Schur's test. Let $A=[a_{ij}]_{i,j\in\mathbb{N}}$ be an infinite matrix. Suppose that there exist positive numbers $p_i>0$, $q_j>0$ ($i,j\in\mathbb{N}$), $\beta>0$ and $\gamma>0$ ...

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First of all you should understand that what is meant when one says that "$\textit{Not all normed linear spaces are inner product spaces.}$" Here is the explanation. An $\textit{inner product space}$ is a vector space with an inner product defined on it. An inner product on $X$ defines a norm on $X$ given by \|x\| = \sqrt{\langle x,x ...

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HINT: Try to prove the following: (1) If $T$ is invertible then so is $T^*$ and $(T^*)^{-1} = (T^{-1})^*$. (2) If $T$ is invertible, then $T^{-1}$ is a bounded linear operator (Open Mapping Theorem).

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