I have some trouble understanding a passage in the proof of the following theorem.

Let $X$ be a normed space and $W \subset X$ a proper subspace of $X$. Let $x \in X$ be such that $d(x, W) = \delta > 0$. Then there exists a functional $f \in X'$, $\|f\| = 1$, such that $f(x) = \delta$ and $f_{|W} = 0$.

Without loss of generality, we can assume $W$ closed. Let $Y = \operatorname{span}\{x\} \oplus W$ and $$\begin{aligned} f \colon Y &\to \mathbb F\\ \alpha x + w &\mapsto \alpha \delta,\qquad w \in W, \alpha \in \mathbb F \end{aligned}$$ It's easy to prove that $f$ is linear, $f_{|W} = 0$ and $f(x) = \delta$. Then $\|f\| \leq 1$: $$|f(\alpha x + w)| = |\alpha\delta| \leq |\alpha| d(x, -\alpha^{-1} w) = \|\alpha x + w\|$$

To prove the remaining part, we apply Riesz's lemma to the spaces $Y$ and $W$: $$\forall \varepsilon \in (0, 1), \exists y_\varepsilon \in Y, \|y_\varepsilon\| = 1 \text{ s.t. } d(y_\varepsilon, W) > 1 - \varepsilon$$ For some $w_\varepsilon, \alpha_\varepsilon$ we have $y_\varepsilon = w_\varepsilon + \alpha_\varepsilon x$. Now for $w \in W$ we have $$\begin{aligned} 1 - \varepsilon &< \|y_\varepsilon - w\| = \|w_\varepsilon - w + \alpha_\varepsilon x\| =\\ &= |\alpha_\varepsilon|\cdot\|\alpha_\varepsilon^{-1}(w_\varepsilon - w) + x\| <\\ &< |\alpha_\varepsilon|\delta(1 + \varepsilon) \end{aligned}$$ where the last inequality follows from the fact that $\alpha_\varepsilon^{-1}(w_\varepsilon - w)$ is an arbitrary element of $W$.

Since $\varepsilon$ was arbitrary, we have that $\|f\| \geq 1$ and the result follows.

In bold there's the part that I fail to understand. If $\delta$ is an infimum, we have that $$\|\alpha_\varepsilon^{-1}(w_\varepsilon - w) + x\| \geq \delta$$ and not the other way around. So, how can we obtain $\delta(1 + \varepsilon)$ as upper bound? It seems to me that we can construct an appropriate $W$ and take a specific $x$ such that $\delta(1 + \varepsilon)$ is not sufficient as an upper bound.


It's not very clearly written. Of course $\delta(1+\varepsilon)$ is not an upper bound, unless $W = \{0\}$, for nontrivial subspaces are unbounded. The point is that there is some $w\in W$ such that we have

$$\lVert \alpha_{\varepsilon}^{-1}(w_{\varepsilon} - w) + x \rVert < \delta(1+\varepsilon),$$

which follows from the definition of $d(x,W)$ since

$$w \mapsto \alpha_{\varepsilon}^{-1}(w_{\varepsilon} - w)$$

is a bijection of $W$.

So we have found $y_{\varepsilon}$ with $\lVert y_{\varepsilon}\rVert = 1$ and

$$\lvert f(y_{\varepsilon})\rvert = \lvert \alpha_{\varepsilon}\rvert\delta > \frac{1-\varepsilon}{1+\varepsilon},$$

whence $\lVert f\rVert > \dfrac{1-\varepsilon}{1+\varepsilon}$.

  • $\begingroup$ Ahh that clears it up. Thank you Daniel, this is perfect! $\endgroup$ – rubik Mar 1 '17 at 13:43

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.