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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. –  user31373 Jul 17 '12 at 15:53
    
You need to do a little work here. @LeonidKovalev's hint makes it almost trivial. –  copper.hat Jul 17 '12 at 16:04
    
@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. –  KiaSure Jul 17 '12 at 18:47
    
You got it, KiaSure. –  Cameron Buie Jul 17 '12 at 19:36
    
@CameronBuie I'm not so sore. KiaSure: see the answer below. –  user31373 Jul 17 '12 at 20:12
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up vote 3 down vote accepted

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 $\|Tx\|_{\ell^p}\le \sum_j \|T_jx\|_{\ell_p}$.

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