Why is $\frac {1}{\lVert f\rVert} = \inf\{\lVert x\rVert : f(x)=1, x \in X\}$ for a linear, bounded functional f?

Question

Let X, Y be normed spaces, f: X $\to$ Y a linear and bounded functional (f non-trival). How can I show that: $$\frac {1}{\lVert f\rVert} = \mathbb{inf} \{\lVert x\rVert : f(x)=1, x \in X\}$$ ? And what do I know from $f(x)=1$? Can I assume that $\lVert f(x)\rVert = 1$

Answer 1

It is clear that $\frac 1 {\|f\|}$ is a lower bound for the set: $1=f(x) \leq \|x\| \|f\|$ so $\frac 1 {\|f\|} \leq \|x\|$.

Now let $\epsilon >0$. There exists $y$ such that $\|y\|=1$ and $f(y) >\|f\|-\epsilon$. Let $x=\frac y {f(y)}$. Then $f(x)=1$ and $\|x\|=\frac 1 {f(y)} <\frac 1 {\|f\|-\epsilon}$. Since $\epsilon$ is arbitrary we are done.