Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $\|Tx\|_{\ell^p}\le \sum_j \|T_jx\|_{\ell_p}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.