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Let $\mathcal{X} =\{ x \in \mathbb{R}^\mathbb{N} : \lim x_n = 0\}$, with the sup norm, and $f: \mathcal{X} \to \mathbb{R}$ be a linear functional defined by $f(x) = \sum_{k = 1}^\infty 2^{-k+1}x_k$. The functional is continuous, with norm $||f|| \leq 2$.

Consider $\alpha \in \mathcal{X} \setminus \ker(f)$. Prove that there is not a point $x \in \ker(f)$ such that $||x - \alpha|| = dist(\alpha, \ker(f))$. I tried using the fact that $dist(\alpha, \ker(f)) = \frac{|f(\alpha)|}{||f||}$ - but to no avail.

Can anyone help me?

user480840
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1 Answers1

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If there were some such $x$, then $\lVert x - \alpha\rVert = \frac{\lVert f(\alpha)\rVert}{\lVert f\rVert}$. We'd have:

$$\begin{align} \lVert f(\alpha)\rVert &= \lVert f(x - \alpha)\rVert \\&\leqslant \lVert f\rVert \lVert x - \alpha\rVert = \lVert f(\alpha)\rVert \end{align}$$

In other words, $\lVert f\rVert = \frac{\lVert f(x - \alpha)\rVert}{\lVert x - \alpha\rVert}$, so the operator norm is attained.

Hint: Is this possible for this operator?

Super Hint:

Remember that $\lVert f \rVert = \sup\{ \lVert f(v)\rVert\,;\, \lVert v \rVert = 1\}$ and consider the sequence $v = (1,1,1,\dots)$.

It does not lie in $\mathcal X$, but it can be approximated by elements of $\mathcal X$. Can any $x\in \mathcal X$ do as good as $v$, with respect to attaining $\lVert f\rVert$?

Fimpellizzeri
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