# Equality in definition of dual space norm

In the definition of the dual space norm, the WP page makes the following statement:

and I was wondering why going from the middle equality to the right equality was obvious?

Basically, because $f$ is linear.
Let's set $$A = \sup\{|f(x)| : x \in X, \|x\| \le 1\}$$ and $$B = \sup\left\{\frac{|f(x)|}{\|x\|}, x \in X, x \ne 0\right\}.$$ Suppose $x \in X$ with $\|x\| \le 1$. If $x=0$ then $f(x) = 0$, so $|f(x)| \le B$. If $x \ne 0$ then $|f(x)|\le \frac{|f(x)|}{\|x\|}$ since the denominator is at most 1. So $|f(x)| \le B$ in either case. Therefore $B$ is an upper bound of the set $\{|f(x)| : x \in X, \|x\|\le 1\}$, so $A \le B$.
Now suppose $x \in X$ with $x \ne 0$. Set $y = \frac{1}{\|x\|} x$. Then by linearity $f(y) = \frac{f(x)}{\|x\|}$, so $|f(y)| = \frac{|f(x)|}{\|x\|}$. Since $\|y\| = 1$, we have $\frac{|f(x)|}{\|x\|} = |f(y)| \le A$. Thus we have $B \le A$.