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

Question

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.

Answer 1

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.