# Verify a given SVD of an operator

Show that the Singular Value Decomposition of the operator

$$A\colon L^2([0,1])\to L^2([0,1]), x\mapsto\int\limits_0^t x(s)\, ds$$

is given by

$$\sigma_j=\frac{1}{(j-1/2)\pi},~~~~~v_j(x)=\sqrt{2}\cos((j-1/2)\pi x),~~~~~u_j(x)=\sqrt{2}\sin((j-1/2)\pi x).$$

My first question is: Is this operator compact at all so that it makes sense to talk of a SVD?

Then: What do I have to do to solve the task?

First, I determined the adjoint operator $A^*$; it is given by $x\mapsto\int\limits_t^1 x(s)\, ds$. Then I attested that $Av_j=\sigma_ju_j$ and $A^*u_j=\sigma_jv_j$.

Moreover, I showed with substitution $\omega=(j-1/2)\pi x$ that

$\langle v_j,v_j\rangle_{L^2([0,1])}=\frac{2}{(j-1/2)\pi}\int\limits_0^{(j-1/2)\pi}\lvert\cos(\omega)\rvert^2\, d\omega=1$ and similarly $\langle u_j,u_j\rangle_{L^2([0,1])}=1$.

Furthermore, I calculated for $k\neq j$ that

$\langle v_j,v_k\rangle_{L^2([0,1])}=\int\limits_0^1 v_j(x)\overline{v_k(x)}\, dx=\int\limits_0^1 v_j(x)v_k(\overline{x})\, dx=\int\limits_0^1 v_j(x)v_k(x)\, dx$ because $x\in [0,1]$. This integral is 0.

Similarly $\langle u_j,u_k\rangle_{L^2([0,1])}=0$.

So far so good. But is this the whole task? Is there something i have to show additionally?

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Concerning compactness, write $(Ax)(t) = \int_0^1 a(t,s)x(s)\,ds$ with $a(t,s) := \chi_{[0,t]}(s)$. Then $\int_0^1 \int_0^1 |a(t,s)|^2 \,ds\,dt < \infty$, hence $A$ defines a Hilbert-Schmidt operator which is compact (see en.wikipedia.org/wiki/Hilbert-Schmidt_integral_operator) – fbg Jan 13 '13 at 13:29