Eigenfunction representation of the L2 derivative I think the main idea of the definitions that follow is to define some sort of generalized double derivative on a subset of $L^2[0,1]$
Define $D(K)$ to be the subset of $C^1[0,1]$ made up of functions $f$ such that $f'(x) = f'(0) + \int_0 ^x \hat{f}(y) dy $ for some $\hat{f} \in L^2[0,1]$, and it is required that $f(0) = f(1)$.
Then we can define the operator $K : D(K) \rightarrow L^2[0,1]$ via $ K: f \mapsto \hat{f}$
The eigenfunctions $f_n(y) = \sqrt{2} \sin(n \pi y)$ form an orthonormal basis of $D(K) \subset L^2[0,1]$
How can I show $K: \sum (f, f_n) f_n \mapsto \sum(-n^2 \pi^2) (f, f_n) f_n$?  (The parentheses means inner product)
I am having trouble dealing with the infinite sum here. $K$ is not a bounded operator so I cannot do the usual thing of passing $K$ through a limit. The first is convergent pointwise since $f$ is differentiable, but the second one is taking place in $L^2$. I know further that $\sum n^4 |(f,f_n)|^2 \iff f \in D(K)$.
 A: A small remark. Note that both series converge naturally in $L^2$, considering that the basis functions are elements of the Hilbert space (and thus, not defined pointwise). We could additionally have almost everywhere pointwise convergence, but this requires proof: it is not direct from Hilbert space rules. 

Answer.
As $f_0 = 0$, there is no constant function in the basis. I'll assume then that we require $f(0)$ and $f(1)$ to be equal to $0$. (This also simplifies the calculations.) 
One way to show it is to analyse each component $(f_n,f)$ of the Fourier series: using integration by parts and the hypotheses on the form of $f'$ we can get somewhere towards a relation with $(f_n, \hat f)$. 
Integrating by parts, we have
$$(f,f_n)=\int_0^1f(y)\sqrt{2}\sin n\pi y\, dy 
= -\int_0^1f'(y)\sqrt{2}\left(-\frac{\cos n\pi y}{n\pi}\right)\, dy
 $$ 
$$= \int_0^1\left( f'(0) + \int_0^y\hat f(t)\, dt\right)\sqrt{2}\left(\frac{\cos n\pi y}{n\pi}\right)\, dy.$$
The term with $f'(0)$ vanishes, so we are left with a double integral. We can apply Fubini's theorem to it and change the order of integration given that the integrand is integrable in absolute value (which in this case is true by the Cauchy-Schwarz-Bunyakovsky inequality). Doing so yields
$$\sqrt{2} \int_0^1\int_0^y\hat f(t)\left(\frac{\cos n\pi y}{n\pi}\right)\, dt\, dy =\sqrt{2} \int_0^1 \hat f(t) \int_t^1\left(\frac{\cos n\pi y}{n\pi}\right)\, dy\, dt $$
$$ = -\sqrt{2} \int_0^1 \hat f(t) \left(\frac{\sin n\pi t}{n^2\pi^2}\right)\, dt = -\frac 1{n^2\pi^2}(\hat f, f_n).$$
Therefore, we get $$(Kf, f_n) = (\hat f, f_n) = -n^2\pi
^2(f,f_n).$$ The convergence in $L^2$ of $Kf:=\hat f$ is guaranteed by hypotheses, so we can say that $\sum n^4|(f,f_n)|^2$ is finite. Moreover, $Kf = \sum (Kf,f_n)f_n = -\sum n^2\pi
^2(f,f_n) f_n$.
