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Consider the real line with Lebesgue measure and $f$, $g$ the characteristic functions of the interval $[0,x]$ for a fixed $x$. Since $\lVert f\rVert_p= x^{1/p}$ and similarly for the $q$ norm, the Hölder's inequality would read $x\leqslant x^{1/p+1/q}$. If $p=q=1$, just pick some $x$ such that $x\gt x^2$. This proves more in the general case: if ...

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For $f_{\alpha,c}(x) = x^{-\alpha}$ on $[c,\infty)$ and $0$ otherwise (and $\alpha > 1$) you have $$\|f_{\alpha,c}\|_1 = \int_{\mathbb{R}} f(x) \,dx = \frac{x^{-\alpha+1}}{1-\alpha}\bigg|_{x=c}^{c=\infty} = \frac{c^{1-\alpha}}{\alpha - 1}$$ and $$\|f_{\alpha,c}\cdot f_{\alpha,c}\|_1 = \|f_{2\alpha,c}\|_1 = \frac{c^{1-2\alpha}}{2\alpha - 1} \text{,} ... 2 Given a Cauchy sequence x_n it is straightforward to see that x_n(i)  converges to some x(i) for all i. Let \epsilon>0 and choose N large enough so that if n,m \ge N, then \|x_n-x_m\|_1 < {\epsilon \over 2}. \begin{eqnarray} \sum_{i=1}^L |x_n(i)-x(i)| &\le & \sum_{i=1}^L |x_n(i)-x_m(i)| + \sum_{i=1}^L |x_m(i)-x(i)|\\ &\le ... 2 It is mostly correct now, but could be written better. Consider l:X'/U^\perp\to U', which sends each coset of U^\perp to its restriction to U. This map is well-defined because all elements of the coset agree on U. The map is surjective, because every functional f on U can be extended to a functional g on X (by Hahn-Banach), and ... 2 One can significantly simplify your proof. Assume B(X,Y) is Banach. The normed space Y can be isometrically be embedded into the B(X,Y) via the map$$ I:Y\mapsto B(X,Y):y\mapsto f(\cdot) y $$for some f\in X^* of norm 1. Since embedding is isometric, then \operatorname{Im} I is a closed subspace of Banach space B(X,Y). Hence \operatorname{Im} ... 2 Here is a sketch of the first part. For each f\in C[0,1], \begin{eqnarray*} \|Tf\| &=& \max_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)f(s)ds\right|\\ &\leq& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)f(s)\right|ds\\ &\leq& \|f\|_{\infty}\max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\ \end{eqnarray*} Therefore \|T\| is ... 2 Fix \varepsilon\gt 0: there exists an integer N such that if n,m\geqslant N and F\subset A is finite, then$$\sum_{\alpha\in F}|f_n(\alpha)-f_m(\alpha)|^2\leqslant\varepsilon. What is important here is that $N$ depends only on $\varepsilon$ but no on the finite set $F$ we are considering. We thus obtain, taking the limit $m \to \infty$, that for ...

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No. You assume that $u_n\overset{w^*}{\to} u$ in $A$ and then asks if $u\in A$. Of course, $u\in A$, because of the assumption. Yes. See this answer. No. Carefully read the proof given in the link.

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My question: I can not show that when $\oplus_\infty X_i$ is separable then $I$ is finite. Literally, that need not be true. The set of $i$ such that $X_i \neq \{0\}$ must be finite. Now, if $J = \left\{i \in I : X_i \neq \{0\}\right\}$ is infinite, then take a sequence $(j_n)_{n\in\mathbb{N}}$ of distinct elements of $J$, for each $j_n$ choose an ...

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$A$ is closed, but not open. The sequence $x=(1,1,1,...)$ is not an interior point. If $(x_n)$ is a convergent sequence in $A$ with the limit, then it also converges pointwise. But since $0\le x_{n,m}\le 1$, we also have $0\le \lim \limits_{n\to \infty} x_{n,m}\le1$, so $\lim \limits_{n\to \infty}x_n\in A.$ $B$ is open, but not closed. If $f\in B$, not that ...

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The fact that dual to dual norm is equal to the original norm in case of finite-dimensional spaces is equivalent to the fact that the corresponding Banach space is reflexive. By James' theorem, a Banach space $B$ is reflexive if and only if every continuous linear functional on $B$ attains its maximum on the closed unit ball in $B$. That is surely true for ...

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