Problem
Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function.
If $f$ has local extrema at $x = a$, then $f(f(x))$ also has local extrema at $x=a$?
The original problem is :
Let $f, g\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function.
If $g$ has local extrema at $x = a$, then $f(g(x))$ also at $x = a$?
This is false, for $f(x)=x^2\displaystyle\sin\left(\frac{1}{x}\right), \ f(0) = 0$ and $g(x)=x^2$
Because, $g$ has local extrema at $x=0$ but, $f(g(x))=f(x^2)$ contains $\displaystyle\sin\left(\frac{1}{x^2}\right)$ which switches its sign infinitely.
And this term makes $f(g(x))$ doesn't have a local extrema at $x=0$
Then how about $f(f(x))$, not $f(g(x))$?
'Infinitely changing sign' occurs this counterexample for $f(g(x))$ case,
But I think the same problem won't be happened for $f(f(x))$. (By the definition of local extrema)
How can I proof this? or is there any counterexample?
You can have the same kind of counterexample for $f \circ f$. Consider for example any differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying $$\forall x \in \left[-1,1 \right], \quad f(x) = x^2 \sin \left( \frac{1}{x}\right) \ \quad \quad \quad \text{and} \quad \quad \quad \forall x \in \left[2,4 \right], \quad f(x) = (x-3)^2 \ $$
Then such an $f$ has a local minimum at $x=3$, whose value is $f(3)=0$, but in a neighbourhood of $x=3$, you will have $$f \circ f (x)= (x-3)^4 \sin \left( \frac{1}{(x-3)^2}\right)$$
which oscillates around $0$ and hence does not have an extremum at $x=3$.