how to find singular values of this operator Could anyone tell me how to find singular values of this operator: $T: L_2(0,1)\to H^1=\{f:  \int_{0}^{1}| f(t)|^2 dt < \infty \text { and } 
\int_{0}^{1}  | f'(t)|^2 dt < \infty\}, T(f)=\int_0^xf(t)dt$, Scalar product in $H^1$ is defined by
$\langle f,g\rangle =    \int_{0}^{1}   f(t)g(t) dt  +  \int_{0}^{1}  f'(t) g'(t) dt $. 
 A: I want consider  the operator $T(g)(x)=\int_0^x g(t)dt-\int_0^1g(t)dt$.
The adjoinit operator of $T$ is the unique map $T^*: H^1\to L^2(0,1)$ such that for every $f\in H^1$ and $g\in L^2(0,1)$ 
$\langle f,T(g) \rangle_{H^1}=\langle T^*(f),g\rangle_{L^2(0,1)}$
$\langle f,T(g) \rangle_{H^1}=\int_0^1 f(x)T(g)(x)dx+\int_0^1 f’(x)T(g)’(x)dx=$
$=\int_0^1 f(x)(\int_0^x g(t)dt)dx-\int_0^1 f(x)(\int_0^1 g(t)dt)dx $
$+\int_0^1f’(x)g(x)dx=$
$=\int_0^1 \frac{d}{dx}(\int_0^x f(t)dt)(\int_0^x g(t)dt)dx -\int_0^1 f(x)(\int_0^1 g(t)dt)dx $
$+\int_0^1 f’(x)g(x)dx=$
$=(\int_0^1 f(t)dt)(\int_0^1 g(t)dt)-$
$-\int_0^1 f(x)(\int_0^1 g(t)dt)dx-\int_0^1 [T(f)(x)-f’(x)]g(x)dx=$
$= \int_0^1 [f’(x)-T(f)(x)]g(x)dx=\langle f’-T(f),g\rangle_{L^2(0,1)}$
So $T^*=D-T$ where $D:H^1\to L_2(0,1)$ is the weak derivative operator. 
Now you must calculate the eighevalues of $T^*T=DT-T^2=I-T^2$
that is self-adjoint and non-negative. 
If $\lambda \in \sigma_p(T^*T)\subset \mathbb{R}^+$ than you have that 
$(I-T^2)-\lambda I=(1-\lambda)I-T^2$
is not injective so there exists $\psi\neq 0$ such that 
$T^2\psi=(1-\lambda)\psi$ so $(1-\lambda)\psi’’=\psi$
If $\lambda=1$ then $\psi=0$ and it is not possible so $ \lambda\neq 1$ and you must resolve 
$\psi’’=\frac{1}{1-\lambda}\psi$
If $0\leq\lambda<1$ then a non zero solution is $ \psi(x)=e^{\frac{x}{\sqrt{1-\lambda}}}$
So $[0,1)\subset\sigma_p(T^*T)$
