# Integrating a $L^{2}$ function in two variables

Consider for $T\in\mathbb{R}_{>0}$ the mapping $$G:\begin{cases} L^{2}((0,T)\times(0,1);\mathbb{R})&\rightarrow X\\ f&\mapsto \int_{0}^{t} f(s,h(s,t,x))\;\mathrm{d} s \end{cases}$$ for a given function $h:[0,T]\times[0,T]\times [0,1]\rightarrow [0,1]$ as smooth as you want. Obviously, we have $$X\subset L^{2}((0,T)\times (0,1);\mathbb{R}),$$ but due to the smoothing of the integral we would hope for a more regular $G(f)$. Can we have a set $X$ where this $G(f)$ belongs to for every $f\in L^{2}((0,T)\times(0,1);\mathbb{R})$? Since the integration is "simultaneously" in the two components the question does not seem to be straight forward.

This brings us to the related question: How smooth must $f$ be such that we obtain $G(f)\in L^{2}((0,T)\times(0,1))$? And of course we could just pick $f\in L^{2}$ as well but we would like to get the most general setting...

Any help or hint is highly appreciated, thank you very much for reading,

Alex

EDIT: Thanks to zhw.'s comment it seems I was too optimistic regarding on how to present my question. Let us make another attempt. Consider the mapping $$\mathcal{G}:\begin{cases} L^{1}((0,T)\times(0,1))&\rightarrow L^{1}((0,T)\times (0,1))\\ f&\mapsto \begin{pmatrix} \int_{0}^{t}f(y,y-t+x)\ \mathrm{d} y & \text{for }\ x\geq t\\ \int_{t-x}^{t}f(y,y-t+x)\ \mathrm{d} y& \text{for }\ x\leq t \end{pmatrix} \end{cases}$$ Then, $\mathcal{G}$ should be well defined. I would like to know if $\mathcal{G}(f)$ for given $f$ is more regular than just $L^{1}$. Can we expect some weak differentiability, maybe some fractional weak derivative at least in one component?

• I don't understand the question. First, you have $X$ at the beginning with no definition. What is it - the range of $G?$ And what is this "Can we have a set" business?
– zhw.
Sep 8 '16 at 20:18

Why is $Gf$ even well defined? Take $T=1.$ Let $h(s,t,x) = s^4.$ Define $f(s,t) = (st)^{-1/4}.$ Then $f\in L^2([0,1]^2).$ But for any $t \in (0,1),$
$$Gf(t,x) = \int_0^t f(s,h(s,t,x))\,ds = \int_0^t f(s,s^4)\,ds = \int_0^t s^{-5/4}\,ds = \infty.$$
• Thank you so much, this is true, I was too optimistic concerning what I need as conditions on $h$. Sorry for that inaccuracy, I will edit the question accordingly.