I have this theorem:
Let X and Y be normed linear spaces and let $T:X\rightarrow Y$ be a linear transformation. The following are equivalent:
a. T is uniformly continuous.
b. T is continuous.
c. T is continuous at 0.
d. there exists a positive real number k such that $\|T(x)\|\le k$ whenever $x \in X$ and $\|x\|\le 1$.
e. there exists a positive real number k such that $\|T(x)\|\le k \|x\|$ for all $x \in X$.
It then says in the book that a consequence of this theorem is:
$\sup\{\|T(x)\|: \|x\| \le 1\}=\inf\{k: \|T(x)\| \le k\|x\|\ \ \forall x \in X\}$
However, I am only able to prove the inequality, not equality:
$\sup\{\|T(x)\|: \|x\| \le 1\}\le\inf\{k: \|T(x)\| \le k\|x\|\ \ \forall x \in X\}$
I get this inequality like this:
Let: $A=\{\|T(x)\|: \|x\| \le 1\}$, $B=\{k: \|T(x)\| \le k\|x\|\ \ \forall x \in X\}$.
Then if $\|T(x^*)\|$ is any element in A, and k is any element in B. Then $\|T(x^*)\|\le k\|x\|\le k$, since $\|x\|\le 1.$
So any element in A is smaller than any element in B.
Then for any a in A, we have that $a \le k$ for any k in B. So a is a lower bound for B, it must be smaller than the biggest lower bound for B, so $a \le \inf B$. But since this holds for any a in A, we must have that $\inf B$ is an upper bound for A, so it must be bigger than the smallest upper bound for A, hence $\sup A\le \inf B$.
Hence I am only able to prove that $\sup A\le \inf B$, but how do I get that $\sup A=\inf B$?
Let $\alpha = \sup(A), \beta = \inf(B)$, and $\epsilon > 0$, then $\beta-\epsilon \notin B$. Hence, $\exists x_0 \in X$ such that $$ \|T(x_0)\| > (\beta-\epsilon)\|x_0\| $$ But then $x_0\neq 0$, so with $y_0 := x_0/\|x_0\|$, we have $\|y_0\| \leq 1$ and thus $$ \alpha \geq \|T(y_0)\| > \beta-\epsilon $$ This is true for every $\epsilon > 0$, so $\alpha \geq \beta$