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I have been reviewing uniform convergence of series and improper integrals with parameters. Working through examples I found it easier to prove uniform convergence (using M-test, Dirichlet-Abel test, etc.) than to refute it. I would like to get a better idea of techniques to prove that uniform convergence fails.

I could show that the improper integral $I(r) =\int_0^\infty \frac{\sin(rx)}{x} \, dx$ converges uniformly for $r \geq c > 0$. The Dirichlet test works because $1/x \to 0$ and we have uniformly bounded

$$\left|\int_0^t \sin(rx)\, dx\right| = \frac{|1 - \cos(rt)|}{r} \leq \frac{2}{c}$$

Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous and the improper integral cannot converge uniformly for $r \in [0,\infty)$. In this case we have the information about the value to apply the continuity theorem.

But how can you show the convergence is not uniform directly from the definition of uniform convergence?

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  • $\begingroup$ "Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$ $\endgroup$ – zhw. Jan 19 at 18:00
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We know

$$I=\int_0^\infty \frac{\sin x}{x}\, dx >0.$$

Suppose there exists $M>0$ such that $b>M$ implies

$$|\int_b^\infty \frac{\sin (rx)}{x}\, dx |<\frac{I}{2}$$

for all $r\ge 0.$ Letting $x=y/r$ then gives

$$|\int_{rb}^\infty \frac{\sin y}{y})\, dy |<\frac{I}{2}$$

for all $r\ge 0.$ Now let $r\to 0^+$ to arrive at $I\le I/2,$ contradiction.

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  • $\begingroup$ Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful. $\endgroup$ – scobaco Jan 25 at 20:54

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