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4

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 ...


4

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 ...


1

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 ...


1

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 $$ ...


1

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 $\gamma>0$ ...


1

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 \begin{equation} \|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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