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Help me please with this question.

Let's $Y$ be Banach space, $Z$ - Normed vector space and $(T_{n})_{\mathbb{N}}$ - the sequence in $B(Y,Z)$ so that all sequence $(y_{n})_{\mathbb{N}}$ in Y holds:

if $\left \| y_n \right \|_{n \to \infty }\rightarrow 0$ then $\left \| T_ny_n \right \|_{n \to \infty }\rightarrow 0$.

Prove that: $ \underset{\mathbb{N}}{\sup}\left \| T_n \right \|< \infty $

Thanks!

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1 Answer 1

up vote 4 down vote accepted

Suppose not: then we can find a subsequence $\{T_{n_k}\}$ such that $\lVert T_{n_k}\rVert\geq 2k$. Pick $x_{n_k}$ of norm $1$ such that $\lVert T_{n_k}(x_{n_k})\rVert\geq \frac{\lVert T_{n_k}\rVert}2$. Then the sequence $\{y_{n_k}:=\frac 1{\lVert T_{n_k}\rVert}x_{n_k}\}$ converges to $0$. But $$\lVert Ty_{n_k}\rVert =\lVert T_{n_k}\left(\frac 1{\lVert T_{n_k}\rVert}x_{n_k}\right)\rVert=\frac 1{\lVert T_{n_k}\rVert}\lVert T_{n_k}(x_{n_k})\rVert\geq\frac 12,$$ a contradiction.


Note that we don't need $Y$ to be a Banach space.

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How do you prove the existence of a $y_{n_k}$ with norm 1 since for large $n$ we have $\|y_n\|<1$? –  Mercy Jun 23 '12 at 10:55
    
$y_n$ is not a good notation, I will change it. Done now. –  Davide Giraudo Jun 23 '12 at 11:05
    
Thanks for your answer! –  Mushka Jun 24 '12 at 10:09
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