Suppose $f=f(x,t)$ is defined on the region $D:=A \times [c,\infty)\subseteq \mathbb R^2,$ and suppose $$\int_c^{\infty} f(x,t)dt$$ exists for all $x \in A.$ Call this improper integral uniformly Cauchy if, for all $\epsilon >0,$ there exists $M>c$ such that $$\left \lvert \int_u^v f(x,t)dt \right \rvert< \epsilon$$ whenever $u,v \geq M$ and $x \in A.$

I have been able to show that uniformly convergent $\implies$ uniformly Cauchy, and I am wondering if the converse is true. I think I was able to show that it is, but I am not entirely convinced by my proof. For brevity I will omit the details, but essentially I tried to show that if the improper integral is uniformly Cauchy, then the sequence of functions defined by $$F_n(x):=\int_c^{c+n} f(x,t)dt$$ is uniformly Cauchy, and therefore converges uniformly to $F(x):=\lim_{n\to \infty} F_n(x).$ Then I think I was able to show that $\int_c^{\infty} f(x,t)dt$ converges uniformly to $F.$

So essentially my question is, is it even true that uniformly Cauchy $\implies$ uniformly convergent? If so, then does my proof sketch seem correct? Thanks!

  • 1
    $\begingroup$ The converse is true. You have the right idea missing the details. $\endgroup$
    – RRL
    Jul 15 '20 at 0:57

Having shown that $\{F_n(x)\}_n$ is uniformly Cauchy, we can claim there exists a function $I:A \to \mathbb{R}$ such that $F_n(x) \to I(x)$ uniformly on $A$.

We have,

$$\left|\int_c^d f(x,t) \, dt - I(x) \right| \leqslant \underbrace{|F_n(x) - I(x)|}_{\alpha(x)} + \underbrace{\left|\int_c^d f(x,t) \, dt - \int_c^{c+n} f(x,t) \, dt\right|}_{\beta(x)} $$

There exists a positive integer $N(\epsilon)$ such that if $n \geqslant N(\epsilon)$, then $\alpha(x)< \epsilon/2$ for all $x \in A$.

Finally, using the uniform Cauchy condition, show that there exists $C(\epsilon)> c +N(\epsilon)$ (independent of $x$) such that if $n = N(\epsilon)$ and $d > C(\epsilon)$, then

$$\beta(x) = \left|\int_{c+N(\epsilon)}^d f(x,t) \, dt\right|< \epsilon/2,$$

for all $x \in A$.


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