Continuity of pr a function defined by an definite integral

Question

Let $f \in {L^2}({(0,1)^2})$ , we defined $F$ by $$F(x) = \int\limits_0^1 {f(s,x)ds} $$ What can I say about the continuity of $F$?. I think as the following: let $x_0 \in L^2(0,1)$, and let $(x_n)$ be a sequence converging to $x$ in $L^2(0,1)$, by the dominate convergence theorem, we can pass to the limit. Is this correct ? Thanks.

Answer 1

No, your assert is certainly false. The correct assumptions are:

• For every $x\in [0, 1]$ set $g_x(s)=f(s, x)$ we must have $g_x\in L^1([0, 1])$. Now because $[0, 1]$ has finite measure we have $L^2[0, 1]\subseteq L^1[0, 1]$ then your $f$ satisfy this assertion;
• For almost every $s\in [0, 1]$ the function $h_s(x)=f(s, x)$ must be continue on all $[0, 1]$ and $f$ doesn't usually satisfy it.

For example take $f(s, x)$ such that $$ f(s, x)=\begin{cases} 0 &\text{ if }x=0\\ 1 &\text{ if }x>0 \end{cases} $$ then $f\in L^2\left([0, 1]^2\right)$ but $$ F(x)=\begin{cases} 0 &\text{ if }x=0\\ 1 &\text{ if }x>0 \end{cases} $$ and $F$ isn't continue.

You need $f$ continue respect to $x$ in order to prove $F$ continuity.