# Uniform convergence of improper integrals with squared integrand

Let the improper integral $\int_0^\infty f(x,t)dx$ converge uniformly on $t \in (0,\infty)$ and suppose $\int_0^\infty [f(x,t)]^2dx$ converges for each $t$. Does the integral $\int_0^\infty [f(x,t)]^2dx$ also converge uniformly?

(1) $f(x,t)$ can be positive and negative. In this case I think it is false. If $\lim_{x \to \infty}\phi(x,t) = 0$ uniformly and montonically then from the Dirichlet theorem the integral $\int_0^\infty \phi(x,t) \sin(x)dx$ converges uniformly because $\int_0^y\sin(x)dx$ is bounded. But $\int_0^y\sin^2(x)dx$ is not bounded and $\int_0^\infty |\phi(x,t)|^2 \sin^2(x)dx$ may not converge uniformly.

(2) $f(x,t) >0$. In this case I think it may be true but I'm having difficulty proving it.

• Forget about uniform convergence, ask first if the convergence of $\int_0^{\infty} f(x)\, dx$ implies convergence of $\int_0^{\infty} |f(x)|^{2}\, dx$. This is not true even if $f$ is positive. Jun 1 '18 at 6:37
• Sorry - I meant to include the condition the $\int_0^\infty |f(x,t)|^2dx$ converges pointwise. I edited the question. Jun 8 '18 at 22:14

If $$f$$ is nonnegative and uniformly bounded then uniform convergence of the improper integral of $$f$$ implies uniform convergence of the improper integral of $$f^2$$. Indeed, if $$0 \leqslant f(x,t) < B$$ for all $$x,t$$ then

$$\int_{c_1}^{c_2} f^2(x,t) \, dx < B\int_{c_1}^{c_2}f(x,t) \, dx$$

Hence, for any $$\epsilon > 0$$ there exists $$C > 0$$ such that for all $$c_2 > c_1 >c$$ and for all $$t \in (0,\infty)$$ we have

$$\int_{c_1}^{c_2} f^2(x,t) \, dx < B \frac{\epsilon}{B} = \epsilon,$$

implying uniform convergence by the Cauchy criterion.

In general, though, your assumptions do not guarantee the uniform convergence of an improper integral of $$f^2$$ even if it exists (pointwise) for all $$t$$.

For a counterexample, take $$f(x,t) = x^{(t-1)/2} e^{-x/2}$$ for $$(x,t) \in [0,1] \times (0,\infty)$$. Note that the improper integral

$$\int_0^1 x^{\frac{t-1}{2}}e ^{-\frac{x}{2}} \, dx$$

converges uniformly for $$t \in (0,\infty)$$ by the Weierstrass M-test since $$f(x,t) \leqslant x^{-1/2}$$ for all $$x \in (0,1]$$.

However, for each $$t \in (0,\infty)$$ we have

$$\tag{*}\int_0^1 f^2(x,t) \, dx = \int_0^1 x^{t-1}e ^{-x} \, dx < \int_0^\infty x^{t-1}e ^{-x} \, dx = \Gamma(t) < \infty,$$

but the convergence is not uniform. To see this, note that for $$t_n < 1$$,

$$\left| \int_{1/2n}^{1/n} x^{t_n-1} e^{-x} \, dx\right| > e^{-1} \left(\frac{1}{n}\right)^{t_n-1}\frac{1}{2n} = \frac{1}{2en^{t_n}}.$$

Taking $$t_n = 1/n \in (0,\infty)$$, we have $$t_n \to 0$$ and $$n^{t_n} = n^{1/n} \to 1$$. For all sufficiently large $$n$$, we have $$n^{1/n} < 2$$ and

$$\left| \int_{1/2n}^{1/n} x^{s_n-1} e^{-x} \, dx\right| > \frac{1}{4e}$$

This violates the Cauchy criterion and the convergence of (*) is not uniform for $$t \in (0,\infty)$$.