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4

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 ...


2

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 ...


1

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.


1

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 ...


1

$A$ 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 ...


1

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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