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## Hot answers tagged functional-analysis

4

Assuming that the operator norm is taken with respect to the Euclidean norm on $\mathbb{R}^n$, we have $$\left\lVert \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}\right\rVert = 1 = \left\lVert \begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\right\rVert,$$ hence $$2\left\lVert\begin{pmatrix}1 & 0\\0&0 \end{pmatrix}\right\rVert^2 + 2 ... 2 This follows directly from the Orlicz–Pettis theorem (applied to the p^{{\rm th}} power of your expression). You can however avoid using this theorem by showing that the function$$f(\phi) = \left( \sum_{n=1}^{\infty} \lvert \phi(x(n))\rvert^p \right)^{1/p}\quad (\phi\in B_{E^*})$$is continuous with respect to the weak*-topology. Then it will be ... 2 First we prove the theorem for the case when \alpha=1/2 and \beta=1/2. In this case, we have$$Q^n (P_1-P_2)=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}P_1^kP_2^{n-k}(P_1-P_2)\\=\frac{1}{2^n}\sum_{k=1}^n[\binom{n}{k-1}-\binom{n}{k}]P_1^kP_2^{n-k-1}+\frac{1}{2^n}P_1^{n+1}-\frac{1}{2^n}P_2^{n+1}$$thus we have that$$||Q^n (P_1-P_2)|| \leq ...

1

The part of the proof that you apparently understand (before "Now it is claimed" in your question) shows that every $y$ with $\Vert y\Vert<\eta$ is the limit of a sequence $Ax_i$ in the range of $A$, with $\Vert x_i\Vert\leq2k$ for all $i$. Now suppose you want to approximate some $y$ whose norm is not known to be $<\eta$. Well, consider ...

1

Any normed space is also a vector space. However, you do not need to define a norm on a vector space, meaning that all vector spaces are a union of the "normed space" rectangle and another rectangle that is outside the big "topological space" rectangle.

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