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Define $\ell^p = \{ x (= \{ x_n \}_{-\infty}^\infty) \;\; | \;\; \| x \|_{\ell^p} < \infty \} $ with $\| x \|_{\ell^p} = ( \sum_{n=-\infty}^\infty \|x_n \|^p )^{1/p} $ if $ 1 \leqslant p <\infty $, and $ \| x \|_{\ell^p} = \sup _{n} | x_n | $ if $ p = \infty $. Let $k = \{ k_n \}_{-\infty}^\infty \in \ell^1 $.

Now define the operator $T$ , for $x \in \ell^p$ , $$ (Tx)_n = \sum_{j=-\infty}^\infty k_{n-j} x_j \;\;(n \in \mathbb Z).$$ Then prove that $T\colon\ell^p \to\ell^p$ is a bounded, linear operator with $$ \| Tx \|_{\ell^p} \leqslant \| k \|_{\ell^1} \| x \|_{\ell^p}. $$

Would you give me a proof for this problem?

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    No, I will not. But here is a hint: for a fixed $j$, consider the simple operator $x\mapsto (k_{j}x_{n-j})$ (sequence indexed by $n$) and find its norm.2012-07-17
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    You need to do a little work here. @LeonidKovalev's hint makes it almost trivial.2012-07-17
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    @LeonidKovalev Thenk you Leonid Kovalev, but $ \sum_n \sum_j |k_{n-j} |^p |x_j |^p = \sum_j \sum_n |k_j|^p |x_{n-j} |^p $ holds? I have a little doubt for this.2012-07-17
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    You got it, KiaSure.2012-07-17
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    @CameronBuie I'm not so sore. KiaSure: see the answer below.2012-07-17
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    @LeonidKovalev: I think it holds (though I don't know how helpful it is). First, put $m=n-j$, so that $\sum_n\sum_j|k_{n-j}|^p|x_j|^p=\sum_n\sum_m|k_m|^p|x_{n-m}|^p$, then just swap the order of the sums and reindex by $j=m$. Since we're dealing with all non-negative terms, I think that's all okay, right?2012-07-17
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    @CameronBuie The index shift is no problem, of course, I was thinking of a different value of "it".2012-07-17

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In the first comment I suggested the following strategy: write $T=\sum_j T_j$, where $T_j$ is a linear operator defined by $T_jx=\{k_jx_{n-j}\}$. You should check that this is indeed correct, i.e., summing $T_j$ over $j$ indeed gives $T$. Next, show that $\|T_j\|=|k_j|$ using the definition of the operator norm. Finally, use the triangle inequality $\|Tx\|_{\ell^p}\le \sum_j \|T_jx\|_{\ell_p}$.