# Proving that $|x^*(x)| \leq \rho(x,Y)$

Exercise :

Let $$(X,\|\cdot\|)$$ be a normed space, $$Y$$ a subspace of $$X$$ and $$x^* \in X$$ with $$\|x^*\| \leq 1$$ such that $$x^*|_Y = 0$$. Show that $$\forall x \in X \setminus Y$$, it is : $$|x^*(x)| \leq \rho(x,Y)$$.

Attempt :

Let $$x \in X \setminus Y$$, thus $$\forall y \in Y$$, it is :

$$|x^*(x)| = |x^*(x) - 0| = |x^*(x) - x^*(y)| = |x^*(x-y)| \leq \|x^*\| \cdot \|x-y\| \leq \|x-y\|$$

Now, there's a hint mentioning that $$|x^*(x)|$$ is the lower bound of $$\{\|x-y\| : y \in Y\}$$ and that the definition of $$\inf$$ shall be used, but I don't understand either why that's the lower bound or what I should do with the infimum.

Any clarification or elaboration on the final part will be much appreciated.

## 1 Answer

By definition $$\rho (x,Y)=\inf \{\|x-y\|:y\in Y\}$$. You have already proved that $$|x^{*}(x)| \leq \|x-y\|$$ for all $$y \in Y$$. Just taking infimum on both sides completes the proof.

• Simple as that it is then ! I always seem to get stuck on the final and easiest parts of such proofs. – Rebellos Nov 11 '18 at 11:56