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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 1 '18 at 6:37
  • 1
    $\begingroup$ Sorry - I meant to include the condition the $\int_0^\infty |f(x,t)|^2dx$ converges pointwise. I edited the question. $\endgroup$
    – scobaco
    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 $t \in (0,\infty)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.