# Uniform boundedness theorem.

Let $$V$$ be subspace of $$\ell^2$$ which contains all 1 summable sequences. For each natural number $$n$$, define $$T_n: V \to \mathbb R$$ by $$T_n(x)=\sum_{i=1}^n x_i$$. Then $$T_n$$ is not uniformly bounded on unit ball $$\|x\|_2\leq1$$.

My intuition says it has something to do with closed and bounded in infinite dimensional banach space need not be compact. But I don't know how to get a firm answer. Could you please tell me the reason? Thank you very much for your time.

• $T_n$ is the sequence of linear functional. – Ziya Oct 11 '18 at 15:38
• Why $T_n$ is not well defined? – Ziya Oct 11 '18 at 15:40
• Do you mean $T_nx=\sum_{k=1}^nx(k)$? (writing elements of $\ell^2$ as functions $x:\mathbb N\to\mathbb C$) – Aweygan Oct 11 '18 at 15:41
• Yes. Actually I am using this site on mobile and there it is not very easy to type it. – Ziya Oct 11 '18 at 15:43

Basically you need to find $$x$$ such that $$\lVert x \rVert_2 \leq 1$$ but $$\lvert T_n(x) \rvert \to \infty$$. Put explicitly you require $$x$$ to satisfy $$\sum_{i=0}^{\infty} x_i^2 \leq 1 \quad \text{and} \quad \sum_{i=1}^{\infty} x_i = \infty$$ Can you think of such $$x$$?
• I tried to find such a $x$ but could not find it. – Ziya Oct 11 '18 at 17:21
• Is it ${\frac{1}{n}}$? – Ziya Oct 11 '18 at 23:59
• Yes! Of course we need to rescale by $\left(\sum_n \frac{1}{n^2}\right)^{-1}$ to get a norm $\leq 1$ but that's not a very important point. – bitesizebo Oct 12 '18 at 0:11