If $f:\mathbb R^2\longrightarrow \mathbb R$ continuous, does $\lim_{h\to 0}\int_0^{\infty } f(t+h,x)dx=\int_0^\infty f(t,x)dx$?

Question

If $f:\mathbb R^2\to \mathbb R$ continuous, s.t. $\int_{\mathbb R}f(t,x)dx$ exist for all $t$. Does $$\lim_{h\to 0}\int_0^{\infty } f(t+h,x)dx=\int_0^\infty f(t,x)dx \ \ ?$$

I asked here the question for $$\lim_{h\to 0}\int_a^b f(t+h,x)dx=\int_a^b f(t,x)dx,$$ where $a,b\in\mathbb R$ and as @Surb show, it's indeed true. But what happen when $a=-\infty $ and $b=\infty $ ? I think it's enough for $a=0$ and $b=\infty $, i.e. does

$$\lim_{h\to 0 }\int_0^\infty f(t+h,x)dx=\int_0^\infty f(t,x)dx \ \ ?$$ (we suppose of course that $\int_0^\infty f(t,x)dx$ exist for all $t$).

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My idea was to show that $$F(v,t):=\int_0^v f(t,x)dx,$$ is continuous on $[0,\infty )\times \mathbb R$, and thus $$\begin{align}\lim_{h\to 0}\int_0^\infty f(t,x)dx=&\lim_{h\to 0}\lim_{v\to \infty }\int_0^v f(t+h,x)dx\\ &=\lim_{v\to \infty }\lim_{h\to 0}\int_0^v f(t+h,x)dx\\ &=\lim_{v\to \infty }\int_0^v f(t,x)dx\\ &=\int_0^\infty f(t,x)dx\end{align}$$

but unfortunately, I didn't success to show that it's continuous. May be it's not correct. And in this case, do you have a counter example ?

Answer 1

Consider $f : \mathbb{R}^2 \to \mathbb{R}$ given by $f(t,x) = t^2e^{-t^2x}$.

We have $$\int_0^\infty f(0,x)\,dx = \int_0^\infty 0\,dx = 0$$

but

$$\lim_{h\to0} \int_0^\infty f(h,x)\,dx = \lim_{h\to0} \int_0^\infty h^2e^{-h^2x}\,dx = \lim_{h\to 0} \left[-e^{-h^2x}\right]_0^\infty = 1$$

Answer 2

This is not true: we will define a function $f(t, x)$ by describing it as a function of $x$ for a "fixed" $t$. For $t = 0$, we set $f(0, x) = 0$. For $t > 0$, we let the function $x \mapsto f(t, x)$ take the value 0 whenever $|x - 1/t| \geq 1$, and we set $f(t, 1/t) = 1$, and we complete the function $x \mapsto f(t, x)$ by making it piecewise linear: starting at $x = 1/t - 1$, the value of $f$ linearly increases to 1 as $x$ goes to $1/t$, and then it linearly descends to $0$ as $x$ goes to $1/t + 1$. Finally, to extend the function to $t < 0$, we set $f(t, x) = f(-t, x)$.

You can check that this function is continuous. Furthermore, trivially for all $t$ with $0 < |t| < 1$ we have $$ \int_0^\infty f(t, x)\,\mathrm dx = 1, $$ so that $$ \lim_{h \to 0}\int_0^\infty f(h, x)\,\mathrm dx = 1 \neq 0 = \int_0^\infty f(0, x) \,\mathrm dx. $$