Exists an $h$ such that $\lVert u + hv \rVert_{\infty}= \max_{x \in U}|u(x)+hv(x)|$

Question

Let $u,v \in C[a,b]$ and $U=\{x\in [a,b]: \lVert u \rVert_{\infty} = |u(x)| \}$. Is it true then that there always exists $0 < h<1$ such that

$$\lVert u + hv \rVert_{\infty}= \max_{x \in U}|u(x)+hv(x)|?$$

Answer 1

Let's try $[a, b] = [0, 1]$ with $u(x) = 1-x$ and $v(x) = \sqrt{x}$. The set $U$ is $\{0\}$, but no matter how small $h\in (0, 1)$ is, the function $u+hv $ is not maximized at $0$, as one can see from the sign of its derivative: $$ (u+hv)'(x) = \frac{h}{2\sqrt{x}} - 1 > 0 \quad \text{ for small $x$} $$