When does $f(a)=f(b)\implies f'(a)=f'(b)$?

Question

My course problem booklet (mathematics BSc, second-year module in algebra, unpublished) has a question,

Let $V=C^\infty(\mathbb{R},\mathbb{R})$, and let $U=\{f \in V: f(x+2\pi)=f(x)\;\forall x \in \mathbb{R}\}$. Show that the linear operator $\frac{\text{d}}{\text{d}x}$ maps the subspace $U$ to itself.

The solution booklet has,

If $f\in U$, $f(x+2\pi)=f(x)$. Differentiating with respect to $x$, $f'(x+2\pi)=f'(x)$, so $f'\in U$, as required.

How does this work? Two functions being equal at a point doesn't guarantee that their derivatives at that point are equal. For instance, if $f_1(x)=e^x, \; f_2(x)=x^2+1,$ then \begin{aligned} f_1(0) & = e^0 \\ & = 1 \\ & = 0^2+1 \\ & = f_2(0), \end{aligned} but \begin{aligned} f_1'(0) & = e^0 \\ & = 1 \\ & \neq0 \\ & = 2 \cdot 0 \\ & = f_2'(0). \end{aligned} I realise that in the problem it's one function, not two, but I don't understand how that gets us there.

Answer 1

Yeah but your identity $f(x+2\pi)=f(x)$ holds for any points, so that the two functions

$f\circ \gamma$ and $f$

are equal on $\mathbb{R}$, where $\gamma(x):= x+2\pi$.

If two functions are equal, of course their derivatives must be equal, and so

$(f\circ \gamma)’(x)=f’(\gamma (x))\gamma’(x)=f’(x+2\pi)=f’(x)$

so that $f’ $ belongs to $U$

Answer 2

As you write down, a function $f$ in $U$ verifies $\forall x \in \mathbb R, f(x+ 2 \pi) = f(x)$. The variable $x$ goes trough the whole real line.

Answer 3

$\lim_{h\to 0}\frac{f(x+2\pi + h)-f(x+2\pi)}{h}=\lim_{h\to 0}\frac{f(x+ h)-f(x)}{h}$