Dual Norm: $\|fx\|\leq \|f\|\|x\|$ ???

Question

I'm studying Functional analysis and at this Wikipedia link: https://en.wikipedia.org/wiki/Dual_norm

In the proof of Theorem 1 there is one passage where apparently it's used the fact that:

$$\|fx\|\leq \|f\|\|x\|$$

I can't understand why this is true.

Answer 1

Do you mean $\|fx\|\leq\|f\|\|x\|$? That's by the definition $$ \|f\| := \sup\{|f(x)| : x\in X,\ \|x\|\leq1\}. $$ It clearly holds if $x=0$, and otherwise consider $\frac{x}{\|x\|}\in X$ and $\|\frac{x}{\|x\|}\| \leq 1$. Then by linearity $$ \|f\| \geq \left|f\left(\frac{x}{\|x\|}\right)\right| = \frac{|f(x)|}{\|x\|} $$ and hence $$ |f(x)|\leq \|f\|\|x\|. $$

Answer 2

$$\|f\|=\sup_{\|x\|=1}\|f(x )\|$$

For any $x \ne 0$, $$\|f\| \ge \|f(\frac{x}{\|x\|})\|=\frac{1}{\|x\|}\|f(x)\|$$ which implies the claim. Obviously the claim holds for $x=0$.