Uniform Convergence and continuity (Rudin 7.4)

Question

This question has been asked for sequences, but I couldn't find it for series.

Suppose $f(x) = \Sigma_{n=1} ^ \infty \frac{1}{1+n^2x}.$

$f(x)$ converges on $(0,\infty)$ and on $(-\infty, 0)$ excluding $\frac{-1}{n^2} $, $\; n = 1,2,3 . . ..$

The question (Rudin 7.4), asks if it is continuous on $(0,\infty)$ and on $(-\infty, 0)$ excluding $\frac{-1}{n^2} $, $\; n = 1,2,3 . . ..$ .

The answer is yes, but I am having trouble proving it. Here's my attempt

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We need to show that given $\epsilon > 0 $ , there is a $\delta$ s.t. $|y-x| < \delta$, implies $|f(y)-f(x)|<\epsilon$.

I tried the $3-\epsilon $ approach.

$$|f(y)-f(x)|< |f_n(y)-f(y)| +|f_n(y)-f_n(x)| + |f_n(x)-f(x)| $$

$$ = | \;\Sigma_{n=1} ^ \infty \frac{1}{1+n^2y} \; - \; \frac{1}{1+n^2y} \;| + | \;\frac{1}{1+n^2y} \; - \; \frac{1}{1+n^2x} \; | $$

$$+| \;\Sigma_{n=1} ^ \infty \frac{1}{1+n^2y} \; - \; \Sigma_{n=1} ^ \infty \frac{1}{1+n^2x} \;| $$

Now, the middle term I can bound becuase of the continuity of the $f_n$'s

but $f$ is a sum, so I don't know if that changes anything? can I say $f_n(x)$ is $\epsilon $ -close to an $f$ given that $f$ is an infinite sum?? Or can I just treat $f$ like a function that $f_n$ converges to uniformly?

Answer 1

A start: Work on $(0,\infty)$ first. Suppose $a>0.$ Claim: the series converges uniformly on $[a,\infty).$ Proof: For $x\in [a,\infty),$

$$0\le\frac{1}{1+n^2x} \le \frac{1}{n^2x}\le \frac{1}{a} \frac{1}{n^2}.$$

Since $\sum \dfrac{1}{a}\dfrac{1}{n^2}<\infty,$ Weierstrass M shows the series converges uniformly on $[a,\infty).$ Since each summand is continuous on $[a,\infty),$ $f$ is continuous on $[a,\infty).$ Now $a>0$ is arbitrary, so it follows that $f$ is continuous on $(0,\infty).$

Answer 2

Let $f_k(x) = \sum_{n=1}^k \frac{1}{1+n^2 x}$ and note that it is continuous everywhere except possibly at $0$ and $-1/n^2$ for $n=1,2,\ldots$.

If you show that $f_k$ converges uniformly to $f$ on some closed interval $I$ that excludes those bad points, then $f$ is continuous on $I$.

To show uniform convergence, you need to bound $$\sup_{x \in I} |f_k(x) - f(x)| \le \sup_{x \in I} \sum_{n=k+1}^\infty \frac{1}{|1+n^2 x|}$$ which I think should not be difficult (the maximizing $x$ will be one of the endpoints of $I$).