The norm of a functional $f$.

Question

I don't understand why $||f||=sup_{||x|| \leq 1} |f(x)|$

Let $f$ be a functional on normed space $X$ which is bounded.

By the definition, $|f(x)| \leq M||x||$. Therefore, $||f|| = inf\{M:|f(x)| \leq M||x||, for~all ~x \in X\}$

Therefore, $||f|| = sup_{x\neq\theta}\frac{|f(x)|}{||x||}$

I can understand it.

However, I don't understand the below expression.

$sup_{x\neq\theta}\frac{|f(x)|}{||x||} = sup_{||x||\leq 1}|f(x)|= sup_{||x||= 1}|f(x)|$

why the $||x|| \leq 1$, $||x||=1$? Definitely, the definition don't say the upper bound of norm.

Answer 1

It can be shown that these expressions $\newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\abs}[1]{\left| #1 \right|}$ $$\begin{align} n_0 &= \sup_{\norm{x} \leq 1} \abs{f(x)}\\ n_1 &= \sup_{\norm{x} = 1} \abs{f(x)}\\ n_2 &= \sup\left\{\frac{\abs{f(x)}}{\norm{x}}\mid x \neq 0 \right\}\\ n_3 &= \inf \{c \geq 0\mid \forall x \in X, \abs{f(x)} \leq c\norm{x} \} \end{align}$$ are equivalent, so $n_0 = n_1 = n_2 = n_3$.

Answer 2

Observe that $$ \{x:x\in X,\|x\|=1\}=\left\{\frac{x}{\|x\|}:x\in X,x\neq0\right\}. $$ Thus, $$ \{|f(x)|:x\in X,\|x\|=1\}=\left\{\frac{|f(x)|}{\|x\|}:x\in X,x\neq0\right\} $$ so we have $$ \sup_{\|x\|=1}|f(x)|=\sup_{x\neq0}\frac{|f(x)|}{\|x\|}. $$ This gives one equality. For the rest of it, observe that $$ \{x:x\in X,\|x\|=1\}\subset\{x:x\in X,\|x\|\leq1\}, $$ so that $$ \sup_{\|x\|=1}|f(x)|\leq\sup_{\|x\|\leq1}|f(x)|. $$ And if $x\in X$ is nonzero with $\|x\|\leq1$, then $$ |f(x)|\leq \frac{|f(x)|}{\|x\|}=\left|f\left(\frac{x}{\|x\|}\right)\right|. $$ This implies that $$ \sup_{\|x\|\leq1}|f(x)|\leq\sup_{\|x\|=1}|f(x)|. $$ Thus the two are equal, and therefore all three expressions are in fact equal.