Norm of functional I'm having little trouble with understanding one proof I saw on mathstack.
Let's consider functional
$$f\colon (C[0,1], \Vert\cdot \Vert_\infty) \ni g\rightarrow g(0) - 7g(\frac12)$$
Calculation $\Vert f\Vert_\infty$ was the following:
The norm of $f$ is equal to $8$ as for the map $h(x)=\vert 4x - 2 \vert -1$ we have $\Vert h \Vert_\infty = 1$ and $f(h)=8$.
And I'm not exactly sure how come it ends the justification. Could you please tell me why with referring to theorems on which this way of thinking is based on ?

I'm going to call the operator $T$ for clarity (instead of $f$). For any $g$ which is continuous on $[0,1]$, we have that $$|Tg|\leq |g(0)|+7|g(1/2)|\leq 8\|g\|_\infty.$$ Thus, $$\sup_{\|g\|_\infty=1}|Tg|=\|T\|\leq 8.$$ The particular function in your post shows that this upper bound can be achieved, and since the operator norm is defined as the least upper bound, the above is actually an equality.
