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No. Let $X=c_0$, so $X^*=\ell^1$. Define $T:\ell^1\to\ell^1$ by $$Tx=T(x_1,\dots)=\left(\sum_{j=1}^\infty x_j,0,0,0,\dots\right).$$ Let $V=\{x\in\ell^1:|x_1|<1\}$. If you go back to the definitions you can verify that $V$ is a weak* neighborhood of the origin but there is no weak* neighborhood of the origin mapped into $V$ by $T$. (Given $\delta>0$ ...

4

The only reason the question seems silly is that you include the answer! A locally convex TVS is one that has a basis at the origin consisting of balanced absorbing convex sets. The reason for the emphasis on "convex" is that that's what distinguishes locally convex TVSs from other TVSs: every TVS has a local base consisting of balanced absorbing sets. ...

3

Suppose each $f_n$ is bounded, i.e. for each $n$, we have $\|f_n\|=\sup_{\|x\|=1}|f_n(x)|<\infty$. Define $T_n:X\to \ell^1(\mathbb N)$ by $x\mapsto \sum_{i=1}^n f_i(x)e_i$, where $\{e_i\}$ is the canonical basis for $\ell^1(\mathbb N)$. If $\|x\|=1$ then $$\sup_n\|T_n x\|=\sum_{i=1}^\infty|f_i(x)|<\infty,$$ so by Banach-Steinhaus we have ...

2

This duality holds if and only if $X^*$ has the Radon-Nikodym property with respect to the Haar measure on $\mathbb{T}$ which is equivalent to having the Radon-Nikodym property with respect to the Lebesgue measure on the unit interval. This is Theorem 1 on p. 98 in J. Diestel and J.J. Uhl, Vector measures. Mathematical Surveys, Vol. 15, Amer. Math. ...

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Let's break things down into steps: $L^p$ is a vector space. (Follows from their definition.) They are normed vector spaces: The $L^p$ norm, by definition, is a finite, nonnegative real number for given $f \in L^p$. 2.1 $\|f\|_p=0$ iff $f=0$ in $L^p$. This follows from the fact that if $f\neq 0$ on a set of positive measure, then $\int |f|^p >0.$ 2.2 ...

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What you did for the first part is correct. For the converse, however, there is a problem. You have to prove that the sequence $\left(M_n\right)_{n\geqslant 1}$ is bounded. Your $K$ depends on $X$, and the task is to prove that $K$ is finite. You can solve the question using the uniform boundedness principle: since $Tx$ belongs to $\ell^\infty$ for each ...

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The idea is to diagonalize, as mentioned earlier, but you have to do it carefully: Let $\{x_n\}$ be a countable dense subset of $X$. Not $\{T_n(x_1)\}$ has a convergent subsequence by Bolzano-Weierstrass, which you index with an increasing sequence $\{s(1,n) : n\in \mathbb{N}\} \subset \mathbb{N}$. Now $\{T_{s(1,n)}(x_2)\}$ has a convergent subsequence, ...

1

No. Let $B=c_0$ and let $(e_n)$ be the standard basis. Let $\alpha_j=1$. Then $||p_n||=1$ but $p_n$ does not converge.

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Here is a solution, sorry that it is rather long, I'm sure a lot can be cut. Let $x_1,x_2 \in [y_1,y_2]$, then since $[x_1,x_2]$ is the convex span of $x_1$ and $x_2$ you have $[x_1,x_2] \subset [y_1,y_2]$. Similarly $y_1,y_2 \in [x_1,x_2]$ implies $[y_1,y_2] \subset [x_1,x_2]$. In the following $\alpha, \tilde \alpha, \beta, \gamma \in [0,1]$. Assume ...

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Yes, it is Lipschitz from $A$ to $X$. You have \begin{align*} \frac{x}{\|x\|} - \frac{y}{\|y\|} &= \frac{\|y\| \, x - \|y\|\,y + \|y\|\,y - \|x\|\,y}{\|x\|\,\|y\|} \\ &= \frac{ x - y}{\|x\|} + \frac{\|y\|- \|x\|}{\|x\|\,\|y\|} \, y. \end{align*} Hence, \begin{align*} \Bigg\| \frac{x}{\|x\|} - \frac{y}{\|y\|} \Bigg\| \le \frac{ \| x - y ...

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There is two different yet equivalent definition of Locally convex spaces : one in which the topology endowed by a family of semi-norms, and one in term of absorbent balanced and convex basis. The equivalence between the two definition is rather long to prove but you can find it in Rudin's Functional Analysis. This might be of interest to you : ...

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