Upper bound for the distance between a point and a subspace

Question

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.

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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.

Answer 1

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