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

Consider a mapping $$T_\lambda: \ell^1 \rightarrow \ell^1\quad T_\lambda f:=\{\lambda_1 f_1,\,\lambda_2 f_2,\lambda_3 f_3,\,\cdots\},$$ where $\lambda_n = 1 - \frac{1}{n}$, $\lambda \in \ell^\infty$.

The operator norm is not attained, which can be shown by Hölder inequality. But I am looking for an alternative proof that is more elementary and avoids from using 'advanced' theories such as Hölder inequality.

Any suggestions?


share|cite|improve this question
The "Hölder inequality" in this case is just the obvious fact that $\sum_j |\lambda_j f_j| \le \sum_j |f_j|$ when all $|\lambda_j| \le 1$. I'd hardly call that "advanced". – Robert Israel Oct 19 '12 at 0:14
@RobertIsrael I know. But the issue is Hölder is beyond the scope of teaching. – newbie Oct 19 '12 at 3:23
up vote 1 down vote accepted

$$ \|T_\lambda f\|_1=\sum_n|\lambda_nf_n|=\sum_n\left(1-\frac1n\right)\,|f_n|. $$ It is easy to see that $\|T_\lambda\|=1$. If $\|T_\lambda f\|_1=\|f\|_1$ for some nonzero $f\in\ell^1$, then $$ \sum_n\left(1-\frac1n\right)\,|f_n|=\sum_j|f_n|. $$ As both series converge (absolutely, being of positive terms), we get $$ \sum_n\frac1n\,|f_n|=0. $$ But since $f\ne0$, $\sum_n\frac1n|f_n|>0$, a contradiction.

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.