Termwise differentiation in Hilbert space I have some question on the following problem. 
Let $f\in \mathcal{C}^1([0,\pi])$ with $f(0)=f(\pi)=0$. Assume $$\int_0^\pi|f(x)|^2dx =1.$$
Show that $$\int_0^\pi|f'(x)|^2dx \geq 1.$$
My solution:
Expand  $f$ in a Fourier sine series: $$f(x) \sim \sum_{k=1}^\infty b_k(f)\sin kx. $$
Since $f\in \mathcal{C}^1([0,\pi])$, $f$ is square integrable on $[0,\pi]$ and so by Parseval's theorem and by assumption, we then have $$1 = \int_0^\pi|f(x)|^2dx = \frac{\pi}{2}\sum_{k=1}^\infty b_k(f)^2.$$
Now expand $f'$ in a Fourier cosine series: $$f(x) \sim \frac{a_0}{2} + \sum_{k=1}^\infty a_k(f')\cos kx. $$
For $k\geq 1$ we have, using the assumption $f(0)=f(\pi)=0$, $$a_k(f') = \frac{2}{\pi}\int_{0}^{\pi} f'(x)\cos kxdx = \frac{2}{\pi}\left[f(x)\cos kx\right]_0^\pi+\frac{2}{\pi}k\int_0^\pi f(x) \sin kxdx = kb_k(f).$$
Since $f'\in \mathcal{C}^0([0,\pi])$, $f$ is square integrable on $[0,\pi]$ and so by Parseval's theorem $$\int_0^\pi|f'(x)|^2dx = \frac{\pi}{4}a_0^2 + \frac{\pi}{2}\sum_{k=1}^\infty k^2b_k(f)^2 \geq \frac{\pi}{2}\sum_{k=1}^\infty b_k(f)^2=1.$$
However, in the hints for the exercise it is suggested that one should indeed expand $f$ as a Fourier sine series, and then differetiate it termwise. However, I'm not sure why we would be allowed to differentiate termwise. The exercise is for a chapter on some concepts of Hilbert spaces, such as convergence in norm ($L_2$ in this case). We can therefore assume that the termwise differentiation refers to convergence in $L_2$ sense. Why would this necessarily we allowed?
Thanks!
 A: In some sense, we can differentiate any $L^2$ convergent series termwise: because $L^2$ convergence implies convergence in the sense of distributions, and differentiation is continuous   on the space of distributions. The catch is that the differentiated series is only guaranteed to convergence in the distributional sense, which is a rather weak mode of convergence, and is not so helpful for dealing with $L^2$ estimates.    
A usual way to justify termwise differentiation is to have uniformly convergent series of derivatives. But we don't have this here, since the Fourier series of $f'$ need not converge uniformly, or even pointwise. 
I can't read the mind of the  author who made that remark, but my guess is that they meant to write "argue that $f'$ has Fourier series formed by the derivatives of the series for $f$" (which is what you did), but  wrote "differentiate termwise" for brevity, then got up to get another donut and never thought about that remark again. Hints for exercises are often not the most carefully composed parts of a book.
