bounded linear operator and norm

Question

If $F:X\rightarrow Y$ is a non zero bounded linear operator and $\alpha\geq 0$. Show that $$\inf\{\|x\|:\|F(x)\|=\alpha\}=\frac{\alpha}{\|F\|}.$$

For all $x\in X$ we have $\|F(x)\|\leq \|F\|\|x\|$.

For $x\in X$ such that $\|F(x)\|=\alpha$ we have $\alpha\leq \|F\|\|x\|$ i.e., $\frac{\alpha}{\|F\|}\leq \|x\|$

So, we have $$\frac{\alpha}{\|F\|}\leq \inf\{\|x\|:\|F(x)\|=\alpha\}.$$

To prove the equality, i search for an element $x\in X$ with $||F(x)||=\alpha$ and $\|x\|=\frac{\alpha}{\|F\|}$

As $F$ is nonzero, there exists $y\in X$ such that $F(y)\neq 0$.

Consider the element $x=\frac{\alpha y}{\|F(y)\|}$. For this, $\|F(x)\|=\alpha$. But this does not give $\|x\|=\frac{\alpha}{\|F\|}$

Help me to complete this.

Answer 1

$$ ||F|| = \sup_{||x||=1}||F(x)|| $$ implies

$$ 1 = \inf\{||x||:||F(x)||=||F||\} $$

and for any $\lambda \gt 0$ $$ \lambda = \inf\{||x||:||F(x)||=\lambda||F||\} $$ now set $\lambda = \frac{\alpha}{||F||}$ to obtain the required result

Answer 2

Note that $$ \underbrace{\inf \{ M \mid \|Fx\| \le M\|x\|, \ \forall x \in X \}}_{:=I_1} \le \underbrace{\inf \{ M \mid \|Fx\| \le M\|x\| \text{ for all } x \in X \text{ such that } \alpha = \|Fx\| \}}_{:=I_2}$$

By definition of inf, for all $\epsilon>0$ there is some $x \in X$ with $\alpha = \|Fx\|$ such that $$(I_2 - \epsilon) \|x\| \le \|F(x)\|=\alpha \implies (I_1 - \epsilon) \le (I_2 - \epsilon) \le \dfrac{\alpha}{\|x\|} \le \dfrac{\alpha}{\inf\{\|x\|:\|F(x)\|=\alpha\}}$$

and since $\epsilon$ is arbitrary we get $I_1 \le \dfrac{\alpha}{\inf\{\|x\|:\|F(x)\|=\alpha\}}$. But $I_1 = \|F\|_{op} $. QED