Understanding the last step in a solution.

Question

Here is the question and its solution in this link:

Do we have $\|f\|_\infty\leq |f(0)|+\|f'\|_\infty$?

I am trying to prove that for any $C^1([0,1],\mathbb{R})$ function $f$ this happen : $\|f\|_\infty\leq |f(0)|+\|f'\|_\infty$

And I read the first solution in the link given above, which is:

Let $x\in [0,1]$, the MVT theorem implies that $f(x)-f(0)=f'(c_x)x, c_x\in (0,x)$ this implies that $|f(x)|\leq |f(0)|+|f'(c_x)|$ and $\|f\|_{\infty}\leq |f(0)|+\|f'\|_{\infty}$.

1-But I did not understand where has the $x$ that was in the inequality preceeding this $|f(x)|\leq |f(0)|+|f'(c_x)|$ gone ?

2-Also, I do not understand how this $|f(x)|\leq |f(0)|+|f'(c_x)|$ implies this $\|f\|_{\infty}\leq |f(0)|+\|f'\|_{\infty}$?

Could anyone explains this for me please?

Answer 1

1.Since $x \in [0,1],$ therefore $|x|\leq 1.$ Hence $$|f(x)| = |f(0)+f'(c_x)\ x| \leq |f(0)| + |f'(c_x)|\ | x |\leq |f(0)| +|f'(c_x)|.$$

2. This follows by the definition of $\|\cdot\|_{\infty}:$

Since $\|f'\|_{\infty} = \sup\{|f'(t)|:t \in [0,1]\},$ therefore $\|f'(t)\|\leq \|f'\|_{\infty}.$ In particular, $\|f'(c_x)\| \leq \|f'\|_{\infty}.$ So we have

$$|f(x)| \leq |f(0)| +|f'(c_x)| \leq |f(0)| + \|f'\|_{\infty}.$$

Since this is true for all $x\in [0,1],$ we obtain (taking supremum) $$\|f\|_{\infty} \leq |f(0)| + \|f'\|_{\infty}.$$