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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?

Thanks.

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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

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$$ \|T_\lambda f\|_1=\sum_n|\lambda_nf_n|=\sum_n\left(1-\frac1n\right)\,|f_n|. $$ 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.

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