Fréchet derivative and local maximum

Question

I'm pretty confused with the idea of local maximum in function spaces. Normally having a null Fréchet derivative is a necessary but not sufficient condition for being a local maximum.

Computing the derivative

let $ f:\mathbb{R} \mapsto \mathbb{R}$ be a continuous function. And lets denote the space of such functions $C_{\mathbb{R,R}}$.

$$ F: C_{\mathbb{R,R}} \mapsto C_{\mathbb{R,R}} $$ $$ F: f \mapsto \sin(f)$$

let us compute its derivative at point f:

$$D_F(f)h = \lim_{t\to 0} {F(f+th) - F(f) \over t} $$

where:

1. $ f \in C_{\mathbb{R,R}} $
2. $ g \in C_{\mathbb{R,R}} $
3. $ t \in \mathbb{R}$

then:

$$D_F(f)h = \lim_{t\to 0} {\sin(f+th) - \sin(f) \over t} $$ $$D_F(f)h = \lim_{t\to 0} {\sin(f)\cos(th)+\cos(f)\sin(th) - \sin(f) \over t} $$ $$D_F(f)h = \lim_{t\to 0} {-h\sin(f) (1 - \cos(th))+h\cos(f)\sin(th) \over th} $$

So using: $$\lim_{x\to 0} {\sin(x)\over x} = 1 $$ $$\lim_{x\to 0} {1-\cos(x)\over x} = 0 $$

it reduces to:

$$D_F(f)h = h\cos(f)$$

Local maximum

we obviously have:

$$ 0 \leq ||F(f)||_\infty \leq 1$$

Hence:

if $||F(f)||_\infty = 1$, then f is a local maximum.

So any function $f$ such that $f(x) \equiv \pi \pmod \pi$ has a solution is a local maximum.

Null Fréchet derivative

$$D_F(f) = \cos(f) = 0 \Rightarrow \exists k \in \mathbb{N}, \forall x \in \mathbb{R}, f(x) = k\pi $$

such constant functions are indeed local maximums, but not the only ones. So instead of getting a superset containing all my local maximums, I get a strict subset of it from the nullity of my Fréchet derivative.

Question

As I'm pretty sure the mathematics I'm taught are right and I'm wrong... Where am I wrong ?

Answer 1

The problem is to maximize $$ \|F(f) \|_\infty. $$ However, the maximum is not differentiable. In order to perform the analysis, the function $f\mapsto\|F(f) \|_\infty$ needs to be differentiable