Prove $\lim_{h \to 0^{+}}\frac{\lVert u +hv \rVert_{\infty} - \lVert u \rVert_{\infty}}{h}=\max_{x \in M}(v\cdot \operatorname{sign}(u))$

Question

I have posted this question before but did not get any answer so im trying again:

Let $u,v \in C[a,b]$ and $M=\{x \in [a,b]: \lVert u \rVert_{\infty} = |u(x)| \}$. The Gâteaux derivative of the supremum norm $\lVert \cdot \rVert_{\infty}$ evaluated at $u$ in the direction of $v$ is:

$$\lim_{h \to 0^{+}}\frac{\lVert u +hv \rVert_{\infty} - \lVert u \rVert_{\infty}}{h}=\max_{x \in M}(v(x)\cdot \operatorname{sign}(u(x)))$$

I'm trying to prove this but do not get anywhere. I do not find any site mentioning this equation either. Any tips? I'm fine with a reference that proves this. Thanks

Answer 1

For some particular $h$, suppose that the max of $u + hv$ occurs at some point $c$ that is not in $M$. Let $b$ be a point of $M$, and let $A = \|u\|_\infty$, to save typing. Then $$ (u+hv)(b) = A + h v(b)\\ (u+hv)(c) = u(c) + hv(c) $$ where $u(c) < A$ (because $c \notin M$). I claim that for small enough $h$, the first of these will be larger than the second (i.e., even if a non-$M$ point like $c$ is the max for some $h$, for small enough $h$ it will not be). Here's the argument:

Let $L = A - u(c) > 0$, Then we have that, for any $h$ now, $$ (u+hv)(b) - (u+hv)(c) = A - u(c) + h (v(b) - v(c)) $$ If we pick $h$ so that $L/2 > |h(v(b) - v(c))|$, then we get that $$ (u+hv)(b) - (u+hv)(c) = L + h (v(b) - v(c)) \ge \frac{L}{2} > 0. $$ So for any point $c \notin M$, we see that for small enough $h$, $$ (u+hv)(c) < (u+hv)(b) $$ Hence the abscissa of maximum value of $u+hv$, as $h$ tends to $0$, must be in the set $M$. For some point $b \in M$, that will have the form $$ (u + hv)(b) = L + hv(b) $$ When we subtract from this the value $\| u\|_\infty = L$, we get a numerator $hv(b)$;dividing by $h$, we get $v(b)$.

Thus the limit must be $v(b)$ for some $b \in M$. Looking back at the expression $$ (u + hv)(b) = L + hv(b) $$ where $b$ ranges over all point of $M$, the largest value this can take occurs where $v(b)$ is largest. In particular, the limit on the left is just $\max{x \in M} v(x)$, as required.

[I've assumed here that $u, v \ge 0$, to avoid dealing with more absolute values, etc.]