Example of a continuous function $f$ of two variables such that $\int_I f(t,x) dt$ converges for all $x$ but does not converge uniformly.

Question

Let $I\subset\mathbb{R}$ be an interval and $x_1,x_2\in\mathbb{R}, x_1<x_2$. I'm looking for an example of a continuous function $f:I\times[x_1,x_2]\rightarrow\mathbb{R}$ such that for all $x\in [x_1,x_2]$ an integral $$\int_I |f(t,x)| dt$$ converges (i.e. it is finite), but the integral $$\int_I f(t,x) dt$$ does not converge uniformly on $[x_1,x_2]$. Assuming the endpoints of $I$ are $a,b\in\overline{\mathbb{R}},a<b$, by uniform convergence I mean $$\lim_{u\to b}\sup_{x\in[x_1,x_2]}\left|\int_c^b f(t,x) dt - \int_c^u f(t,x) dt\right| =0$$ $$\lim_{u\to a}\sup_{x\in[x_1,x_2]}\left|\int_a^c f(t,x) dt - \int_u^c f(t,x) dt\right|=0 $$ where $c\in (a,b)$.

Answer 1

The function $f$ given by

$$f(t,x) = xe^{-tx}$$

is continuous on $[0,\infty) \times [0,1],$ and

$$\int_0^\infty |f(t,x)| \, dt = \begin{cases} 1,& x > 0 \\ 0, &x = 0\end{cases}$$

However the integral of $f$ is not uniformly convergent for $x \in [0,1]$ since

$$\lim_{c \to \infty}\sup_{x \in [0,1]}\int_c^\infty xe^{-tx} \, dt = \lim_{c \to \infty}\sup_{x \in [0.1]}e^{-cx} = 1 \neq 0.$$

Answer 2

Sketch: Let's look at $\mathbb R \times [0,1].$ On $\mathbb R,$ define $g(t) = t(1-t)\chi_{[0,1]}(t).$ Then $g$ is continuous on $\mathbb R.$ On $\mathbb R \times [0,1],$ set $f(t,x) = g(t-1/x), x\in (0,1],$ $f(t,0) = 0.$ Then $f$ is continuous and nonnegative on $\mathbb R \times [0,1].$ For any $u>0,$

$$\sup_{x\in [0,1]}\int_u^\infty f(x,t)\, dt = 1/6,$$

showing the limit in question is $1/6.$