Infinity norm inequality in continuous function space.

Question

There's this inequality that i find intuitive but can't seem to prove it nor unprove it. So for $ f: [0,1] \mapsto \mathbb{R}$ a continuous function and $ L : [0,1] \mapsto \mathbb{R} $ an affine function that satisfies $L(0) = f(0)$ and $L(1) = f(1)$. The norm is defined by $\lVert f \rVert_\infty = \sup_{x\in[0,1]}|f(x)|$. Is the following true?

$$ \lVert f - L \rVert_\infty \le \lVert f \rVert_\infty $$

$f - L$ represents for me a skew transformation of the function f. So it seems natural that it wouldn't surpass the maximum value of f itself.

Answer 1

No, this is not true. For the sake of simplicity, allow me to work on $[-\pi, \pi]$ instead of $[0,1]$. Let $f(x) = \cos x$. Clearly, $\| f \|_\infty = 1$. Notice that, in general, $L(x) = (f(\pi) - f(-\pi))x + f(-\pi)$ so, in this case, $L(x) = \cos (-\pi) = -1$. It follows that $\| f - L \| _\infty \ge 2$ because $(f-L) (0) = 1 - (-1) = 2$, so clearly $2 = \| f-L \|_\infty \nleq \| f \| = 1$.

Answer 2

Take $f$ defined by the graph $(0,0), ({1 \over 2}, -1), (1,1)$. Then $L(x) = x$, $\|f\|_\infty = 1$ and $\|f-L\|_\infty = {3 \over 2}$.