how to find singular values of this operator

Question

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 $.

Answer 1

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 $[0,1)\subset\sigma_p(T^*T)$