3
$\begingroup$

I'd like some help to solve the following question:

Let $(a_i)$ be a sequence in $\mathbb{R}$ such that $\sum_{i=1}^\infty a_ix_i< \infty$ for all $(x_i)\in l_1(\mathbb{R})$. Show that $(a_i)\in l_\infty$.

My attempt. For every $n\in \mathbb{N}$, set $f_n:l_1\rightarrow \mathbb{R}$ as $f_n(x)=\sum_{i=1}^na_ix_i$, where $x=(x_i)$. Clearly, $f_n$ is a linear functional in $l_1$. Moreover, since $$|f_n(x)|\leq \sum_{i=1}^n|a_ix_i|\leq \max_{1\leq i\leq n}|a_i| \cdot \|x\|_1,$$ $f_n$ is bounded. I know if I get $\|f_n\|= \max_{1\leq i\leq n}|a_i|$, I can use the uniform boundedness principle to obtain the result. However, I couldn't find an element $x\in l_1$ such that $$\frac{|f_n(x)|}{\|x\|_1}= \max_{1\leq i\leq n}|a_i|.$$ I'll be grateful for any hint. Thanks in advance!

$\endgroup$
2
$\begingroup$

Pick $k \leqslant n$ such that $\lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert$. Then

$$\lvert f_n(e_k)\rvert = \lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert \cdot \lVert e_k\rVert_1$$

for $e_k$ the $k^{\text{th}}$ standard unit vector, i.e.

$$(e_k)_i = \begin{cases} 1 &\text{if } i = k \\ 0 &\text{if } i \neq k \end{cases}\,.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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