Let $S=\{ a \in \ell^2 \setminus \sum_{n=1}^\infty a_n <\infty \}$ be the subspace of $\ell^2$ of summable sequences over $C$.

Let $T:S \to C$ be the linear functional such that $T(a)=\sum_{n=1}^\infty a_n$

My question is: is $T$ a bounded linear functional?

Thanks for any suggestion.


Consider the sequences $(x_i)_{i=1,2,\dots}$ defined by $$(x_i)_n = \frac{1}{n^{1+1/i}}.$$ The sequences are all summable, because the improper integrals $$\int_{1}^{\infty} \frac{1}{x^{1+1/i}}\, dx$$ are finite, so they all belong to $S$. They are also bounded in $\ell^2$, because $$ \sum_{n=1}^{\infty} (x_i)_n^2 \leq \sum_{n=1}^{\infty} \frac{1}{n^2} \quad i = 1,2,\dots.$$ However, $T(x_i) \to \infty$ as $i\to \infty$, because the sequence $x_i$ converges pointwise to $1/n$, which is not summable.

  • $\begingroup$ my question is if $\lVert T \rVert <\infty$ , is your answer for this? $\endgroup$ – Matey Math Jan 27 '17 at 23:35
  • $\begingroup$ By definition, $\|T\| < \infty$ if and only if $|T(x)| \leq c|x|$ for some $c \geq 0$. This can't be the case here, because we have that $|T(x_i)| \to \infty$ even though $x_i$ are bounded in $\ell^2$. So the answer is no. $\endgroup$ – Roberto Rastapopoulos Jan 27 '17 at 23:37
  • $\begingroup$ ok thanks for this $\endgroup$ – Matey Math Jan 27 '17 at 23:38

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