Supremum of Absolutely Continuous Function

Question

Is it true for $f : [0,1] \rightarrow \mathbb{R}$ an absolutely continuous function we have \begin{align*} \sup_{0 \leq y \leq 1} |yf(0)+(1-y)f(1)| \leq \sup_{0 \leq x \leq 1} |f(x)| \end{align*} ?

I suspect this is true for several cases but I cannot prove for general case. Thank you.

Answer 1

The left-hand side is smaller than $\max\{|f(0)|, |f(1)|\}$.

Answer 2

This function $f$ has maximum value, because of its continuity in compact interval, so it is for $|f|$. Thus $sup_{x\in[0,1]}|f(x)|=max_{x\in [0,1]}|f(x)|\geq |f(1)-f(0)|=max_{y\in [0,1]} |yf(0)+(1-y)f(1)|$, because $\frac {d}{dy}(yf(0)+(1-y)f(1))=f(0)-f(1)$, so the function $|g|(y):=|yf(0)+(1-y)f(1)|, y\in [0,1]$ takes its maximum for $|f(0)-f(1)|$.