If a sequence $h_n \in \mathscr{H}$ has that $\lim\limits_{n \to \infty}(h_n, h) \to 0$ for every $h \in \mathscr{H}$, do we have $\sup_n \lVert h_n \rVert < \infty$?

My thought is that: Let $\lambda_n$ be a linear functional from $\mathscr{H}$ defined by $\lambda_n(h) = (h_n,h)$ for $h \in \mathscr{H}$. Then \begin{align*} \lim\limits_{n \to \infty} \lVert \lambda_n \rVert = \lim\limits_{n \to \infty} \sup_n \{ \lvert (h_n, h) \rvert \} = 0 \implies \lVert \lambda_n \rVert < \infty. \end{align*} Then using the Uniform Bounded Principle for Banach space, there exists $M < \infty$ such that $\sup_n \lVert h_n \rVert = \lVert \lambda_n \rVert < \infty$.

But I'm unclear about my assertion that \begin{align*} \lim\limits_{n \to \infty} \lVert \lambda_n \rVert = \lim\limits_{n \to \infty} \sup_n \{ \lvert (h_n, h) \rvert \} = 0 \implies \lVert \lambda_n \rVert < \infty. \end{align*} What if the sequence $\lambda_n$ starts off to have infinite norms? How should I derive that all the linear functionals are bounded?


First note that $\|h_n\| = \|\lambda_n\|$:

We have $$|\lambda_n(x)| = |\langle h_n, x\rangle| \le \|h_n\|\|x\| \implies \|\lambda_n\| \le \|h_n\|$$ $$\|\lambda_n\| \ge \frac{|\lambda_n(h_n)|}{\|h_n\|} = \|h_n\|$$

To note that $\sup_{n\in\mathbb{N}} \|\lambda_n\| < +\infty$, by the Uniform Boundedness Principle it suffices to check that $\sup_{n\in\mathbb{N}} |\lambda_n(x)| < +\infty$ for all $x \in H$.

We have $|\lambda_n(x)| = |\langle h_n, x\rangle| \xrightarrow{n\to\infty} 0$ so it is in particular bounded (since a convergent sequence is bounded). Hence $\sup_{n\in\mathbb{N}} |\lambda_n(x)| < +\infty$.

Therefore, $\sup_{n\in\mathbb{N}} \|\lambda_n\| < +\infty$ and thus $\sup_{n\in\mathbb{N}} \|h_n\| < +\infty$.

  • 1
    $\begingroup$ Ah! A convergent sequence is bounded indeed! $\endgroup$ – nekodesu Mar 24 '18 at 18:26

You have the good idea, the uniform boundness principle says that:

If for every $h\in H$ the sequence $\lambda_n(h)$ is bounded, then, there exists $M$ such that $sup_n(\|\lambda_n\|)<M$. This implies that $\lambda_n({h_n\over{\|h_n\|}})=(h_n,{h_n\over{\|h_n\|}})=\|h_n\|<M$.

  • $\begingroup$ Thanks. My concern is that $\lim_{n \to \infty} \lVert \lambda_n \rVert$ is $0$ doesn't guarantee that $\lambda_n$ is bounded for all $n$. Then UBP doesn't apply here. $\endgroup$ – nekodesu Mar 24 '18 at 14:20
  • $\begingroup$ UBP applies because $(h_n,h)=\lambda_n(h)$ is bounded for every $h$ and this is the result of $lim_n(h_n,h)=0$. $\endgroup$ – Tsemo Aristide Mar 24 '18 at 14:21
  • $\begingroup$ Yeah this seems intuitively correct but what if the sequence $\lambda_n = \infty$ for finitely many $n$ at the beginning of the sequence? $\endgroup$ – nekodesu Mar 24 '18 at 14:34
  • $\begingroup$ UBP is true.... $\endgroup$ – Tsemo Aristide Mar 24 '18 at 14:39

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