From $||(f_1+f_2)x|| \leq ||f_1|| + ||f_2||\ \forall x \text{ such that } ||x||\leq 1$, how does $||f_1+f_2|| \leq ||f_1|| + ||f_2||$ follow? Also from the wiki article on the dual of a norm (sorry for the second question about this in such a short amount of time):
$X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number
$$||f|| = \text{sup}\{|f(x)|:x\in X, ||x|| \leq 1\}.$$
At some point, they conclude that $$||(f_1+f_2)x|| \leq ||f_1|| + ||f_2||\ \forall x \text{ such that } ||x||\leq 1$$
and say that this implies
$$||f_1+f_2|| \leq ||f_1|| + ||f_2||.$$
How does that follow?
I can see how, for any $x$,
\begin{align}||(f_1+f_2)x|| &= ||\big((f_1+f_2)\frac{x}{||x||}||x||\big)|| \\
&=||\big((f_1+f_2)\frac{x}{||x||}\big)||\cdot ||x|| \\
&\leq (||f_1|| + ||f_2||)\cdot ||x||
\end{align}
but don't see how that helps.
 A: If you have
$$
\|(f_1+f_2)x\|\le \|f_1\| + \|f_2\|
$$
for all $x$ with $\|x\|\le 1$, then you can take the supremum over all such $x$ on the left-hand side to conclude
$$
\|f_1+f_2\| = \sup_{x:\ \|x\|\le 1}\|(f_1+f_2)x\|\le \|f_1\| + \|f_2\|.
$$
A: I take it that you want to confirm that the operator norm is indeed a norm, and in particular that the triangle inequality is satisfied. Following the approach of Kreyzwig, let's show that $$\|T\|=\underbrace{\sup \|Tx\|}_{\|x\|=1, x \neq 0}.$$
Well, this is not so bad, let $\|x\|=a$, and then let $y=\frac{1}{a}x$ for $x \neq 0$. Then $\|y\|=1$, and by linearity, we obtain:
$$\|T\|=\sup\frac{1}{a}\|Tx\|=\sup\|T(\frac{1}{a}x)\|=\sup\|Ty\|$$
where $\|y\|=1$. From this point of view, the result should seem a little more obvious, since we are taking a supremum over the values of functions applied to vectors of unit length. We get that
$$\sup_{\|x\|=1}\|(T_1+T_2)x\|=\sup_{\|x\|=1}\|T_1x+T_2x\| \leq \sup_{\|x\|=1}\|T_1x\|+\sup_{\|x\|=1} \|+T_2x\|.$$
The last inequality follows from basic properties of the supremum.
A: So you have shown that for $C=\|f_1\|+\|f_2\|$, 
$$ \|(f_1+f_2)(x)\|\leq C\|x\|, \ \ \ \forall x\in X.$$  
This implies that the functional $f_1+f_2$ is bounded and that $\|f_1+f_2\|\leq C=\|f_1\|+\|f_2\|$.
