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Repost of this Math.SE question due to a lack of answers (No one was able to help me find the closed form of $U_T$ and $|T|$ after two bounties). I also searched extensively online but couldn't find anything. A reference for the solution suffices.

Define the Volterra integral operator (or numerical differentiation operator) $$ T: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{0}^{x} f(y) dy. $$ Find its polar decomposition $T = U_T | T |$.

First I tried to find $|T| := (T^* T)^{\frac{1}{2}}$. We have $$ T^*: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{x}^{1} f(y) dy. $$ and therefore $$ T^* T: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{x}^{1} \int_{0}^{y} f(z) dz dy. $$ Now I got stuck when finding $\sqrt{T^* T}$. I found out that $T^2 \ne T^* T$ and $(T^*)^2 \ne T^* T$. Can somebody please give me a hint?

I have also tried writing out $T^* T$ more explicitly: denote by $F$ the anti-derivative of $f$ and by $\mathscr{F}$ the antiderivative of $F$. The we have $$ (T^* T f)(x) = \int_{x}^{1} F(y) - F(0) dy = \int_{x}^{1} F(y) dy - (1 - x) F(0) = \mathscr{F}(1) - \mathscr{F}(x) - (1 - x)F(0). $$

Edit. I also know that $T^* T$ is compact and self-adjoint and
$$ (T^{n + 1} f) = \frac{1}{n!} \int_0^x (x - y)^n f(y) dy. $$ holds. Maybe this can help use find a closed form for $(I - T^* T)^n$?

Edit 2. I know that $$ | T | = \sum_{n \in \mathbb N} \sigma_n \langle \cdot, v_n \rangle v_n, $$ where $\sigma_n := \frac{1}{\pi\left(n - \frac{1}{2}\right)}$ are the singular values of $T^* T$ and $(v_n)_{n \in \mathbb N} \subset L^2([0,1])$ a orthonormal basis of $\overline{\text{ran}(T)}$ which fulfils $$ T^* T x = \sum_{n \in \mathbb N} \sigma_n^2 \langle x, v_n \rangle, v_n. $$ The eigenfunctions of $T^* T$ are $f_n(x) = C \cdot \cos(\sigma_n^{-1} x)$ for some $C \in \mathbb{R}$.

Partial answer from Math.SE We have $$\sqrt{T^* T} = I-\sum^{\infty}_{n=1}a_n(I-T^*T)^n,$$ where the $a_n$ are determined by $$1-\sqrt{1-x}=\sum^{\infty}_{n=1}a_nx^n.$$ This yields $\sum_{n = 0}^{\infty} a_n = 1$. The first $a_{2n}$ are $$\left(\frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \frac{7}{256}, \frac{21}{1024}, \frac{33}{2048}, \frac{429}{32768}\right),$$ the odd terms are zero. The numerators, omitting the zeros, are A098597 on OEIS and the denominators (omitting zeros again) are given by $2^{\ell}$, $\ell \in \{1,3,4,7,8,10,11,15\}$, which is this in OEIS.

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  • $\begingroup$ @ChristianRemling The reason I hope there exists an explicit answer is that this exercise is taken from this problem set from LMU Munich wherein it states "Important: the full solution here is to produce the explicit "closed" form of the operators $U_T$ and $|T|$. Giving the canonical decomposition of $|V|$ is not enough, show that in fact $|T|$ is an integral operator and $U_T$ is a unitary operator (with an explicit, "easy" form)." $\endgroup$
    – ViktorStein
    Commented Mar 24, 2020 at 8:45
  • $\begingroup$ I see. In that case, we need a kernel $K(x,t)$ with $\overline{K(t,x)}=K(x,t)$ and $\int_0^1 K(x,s)K(s,t)\, ds = 1-\max \{ x,t\}$ (since this latter expression is the kernel of $T^*T$), and you would then hope to get the right idea by just staring at this long enough (?). $\endgroup$
    – Christian Remling
    Commented Mar 24, 2020 at 16:30
  • $\begingroup$ @ChristianRemling I know a kernel for $T$ is $K(x,t) := \mathbb{1}_{t \le x}(x,t)$, as in $$(T f)(x) = \int_0^1 K(x,t) f(t) dt.$$ Futhermore, $K(t,x)$ is a kernel for $T^*$: $$(T^* f)(x) = \int_0^1 K(t,x) f(t) dt.$$ This is consistent with the kernel-free expression I found for $T^* T$ above: $$ (T^* T f)(x) = T^*\left(\int_0^1 K(x,s) f(s) ds \right) = \int_0^1 K(t,x) \int_0^1 K(x,s) f(s) ds dt $$ Am I on the right track? Because I think that $\overline{K(t,x)} = K(t,x) \ne K(x,t)$ and $$\int_0^1 K(x,s) K(set) ds = \int_x^t dx = t - x \ne 1 - \max(x,t).$$ $\endgroup$
    – ViktorStein
    Commented Mar 24, 2020 at 20:36
  • $\begingroup$ So shouldn't the kernel of $T^* T$ be $\tilde{K}(x,t) := \int_0^1 K(t,x) K(x,s) ds$, as $$(T^* T f)x = \int_0^1 \tilde{K}(x,t) f(t) dt?$$ $\endgroup$
    – ViktorStein
    Commented Mar 24, 2020 at 20:54
  • $\begingroup$ I think my formula $\widetilde{K}(s,t)=1-\max\{ t,s \}$ is fine. I simply changed the order of integration in $(T^*Tf)(x)=\int_x^1 dt\int_0^t ds\, f(s)$ to obtain this. $\endgroup$
    – Christian Remling
    Commented Mar 24, 2020 at 21:10

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