Wirtinger's inequality variation 
If $f \in C^1[0,1]$ with $f'(0) = f(1) = 0$, then$$\|f\|_2\leq\frac2\pi\|f'\|_2.$$

Elaboration:
Assume the Sturm-Liouville operator $A: D \longrightarrow L^2(0,1)$ where the domain is
$$
D = \{f \in C^1[0,1]: f'' \in L^2(0,1), f'(0) = f(1) =0\}
$$ 
and
$$
Af(x) = f''(x)-\lambda f(x), \, \lambda \in \mathbb{R}, \, x \in [0,1]
$$
The eigenfuctions of $A$: $\phi_n(x) = \sqrt{2} \cos\left( \frac{(2n-1)\pi}{2}\right), \, n =1,2,\dots$ is an orthonormal basis of $L^2(0,1)$.
Then for an $f \in C_1[0,1]$ with $f'(0) = f(1) = 0$ we have: 
$$
f(x) = \sum_{n=1}^\infty b_n \sqrt{2} \cos\left( \frac{(2n-1)\pi}{2}x\right), \, n =1,2,\dots
$$
Now it'd be very nice if
$$
f'(x) = \sum_{n=1}^\infty a_n \sqrt{2} \sin\left( \frac{(2n-1)\pi}{2}x\right), \, n =1,2,\dots \tag{$*$}
$$
so that, by integrating both sides
$$
\int_1^x f'(s)\,\mathrm ds = \sum_{n=1}^\infty a_n \sqrt{2} \int_1^x\sin\left( \frac{(2n-1)\pi}{2}s\right)\,\mathrm ds\\
f(x) = \sum_{n=1}^\infty \frac{-2a_n}{\pi(2n-1)}\sqrt{2}\cos\left( \frac{(2n-1)\pi}{2}x\right)
$$
and thus by using Parseval theorem:
$$
\|f\|_2^2 = \sum_{n=1}^\infty \frac{4a^2_n}{\pi^2(2n-1)^2} \leq \frac{4}{\pi^2}\sum_{n=1}^\infty a_n^2 = \frac{4}{\pi^2}\|f'\|^2_2
$$
and therefore:
$$
\|f\|_2^2 \leq \frac{2}{\pi} \|f'\|_2^2
$$
Is equation $(*)$ (or some variation of it) true and why?
In other words, can the Fourier series expansion of $f$ be term by term differentiated and why?

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Note that after changing the domain $D$ to$$
D_1 = \{f \in C^1([0, 1]) \mid f'' \in L^2([0, 1]),\ f'(1) = f(0) = 0\},
$$
the Sturm-Liouville theorem implies that $\{ψ_n(x) \mid n \in \mathbb{N}_+\}$ is also an orthonormal basis of $C^1([0, 1])$, where $ψ_n(x) = \sqrt{2} \sin\left( \dfrac{1}{2} (2n - 1)π x \right)$, thus there exist a sequence of constants $\{a_n\}$ such that$$
f'(x) = \sum_{n = 1}^∞ a_n \sqrt{2} \sin\left( \frac{1}{2} (2n - 1)π x \right)
$$
if $f \in C^2([0, 1])$. But for any $f \in C^1([0, 1])$, there exist a sequence of functions $\{f_n\} \subseteq C^2([0, 1])$ that $f_n'$ uniformly converges to $f'$ and $\lim\limits_{n → ∞} f_n(0) = f(0)$, so this suffices for the proof of the inequality.

Actually, there is an identity:

Proposition: If $f \in C^1([0, 1])$ satisfies $f'(0) = f(1) = 0$, then$$
\frac{4}{π^2} \int_0^1 (f'(x))^2 \,\d x - \int_0^1 (f(x))^2 \,\d x = \int_0^1 \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x.
$$

Proof: For $0 < δ < 1$,\begin{gather*}
\int_0^{1 - δ} \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x\\
{\small= \frac{4}{π^2} \int_0^{1 - δ} (f'(x))^2 \,\d x + \frac{4}{π} \int_0^{1 - δ} f(x) f'(x) \tan\left( \frac{π}{2} x \right) \,\d x + \int_0^{1 - δ} (f(x))^2 \tan^2\left( \frac{π}{2} x \right) \,\d x,}\tag{1}
\end{gather*}
and\begin{align*}
&\peq \int_0^{1 - δ} f(x) f'(x) \tan\left( \frac{π}{2} x \right) \,\d x = \frac{1}{2} \int_0^{1 - δ} \tan\left( \frac{π}{2} x \right) \,\d((f(x))^2)\\
&= \frac{1}{2} \left. (f(x))^2 \tan\left( \frac{π}{2} x \right) \right|_0^{1 - δ} - \frac{π}{4} \int_0^{1 - δ} (f(x))^2 \sec^2\left( \frac{π}{2} x \right) \,\d x\\
&= (f(1 - δ))^2 \tan\left( \frac{π}{2} (1 - δ) \right) - \frac{π}{4} \int_0^{1 - δ} (f(x))^2 \sec^2\left( \frac{π}{2} x \right) \,\d x.
\end{align*}
Since $\tan^2 α - \sec^2 α = -1$, then\begin{gather*}
\small(1) = \frac{4}{π^2} \int_0^{1 - δ} (f'(x))^2 \,\d x - \int_0^{1 - δ} (f(x))^2 \,\d x + \frac{4}{π} (f(1 - δ))^2 \tan\left( \frac{π}{2} (1 - δ) \right).\tag{2}
\end{gather*}
Note that as $δ → 0^+$,$$
f(1 - δ) = \int_{1 - δ}^1 f'(x) \,\d x \sim f'(1) δ,\quad \tan\left( \frac{π}{2} (1 - δ) \right) = \cot\left( \frac{π}{2} δ \right) \sim \frac{2}{πδ},
$$
thus making $δ → 0^+$ in (2) yields$$
\int_0^1 \left( \frac{2}{π} f'(x) + f(x) \tan\left( \frac{π}{2} x \right) \right)^2 \,\d x = \frac{4}{π^2} \int_0^1 (f'(x))^2 \,\d x - \int_0^1 (f(x))^2 \,\d x.
$$
