Uniform or non uniform.

Question

Consider the sequence of functions $f_n:(-1,1) \rightarrow \mathbb{R}$, $f_n(x) = \sum_{k=1}^n \frac{n}{n^2-k^2x}$. Find the function $f(x)$ to which $f_n$ converges pointwise. Determine if this convergence if uniform.

Yes. This looks like an exercise. However, this does not come from any text. I have made this problem studying mode of convergence of Riemann sums to the integral. The model function was $f(x)=\int_0^1 \frac{1}{1-t^2x} d t$ as one can easily see. However, the mode of convergence of the Riemann sum is somewhat sophisticated. Neither $f_n$ nor $f$ are bounded on the open interval. So it is unclear if $f_n$ converges to $f$ near the endpoints. Of course, on any compacta in $(-1,1)$, this sequence converges uniformly since the integrand is uniformly bounded. But if $x$ approaches to $\pm 1$, what would happen?

Answer 1

Fix $a\in(0,1)$. For $x\in[-1,a]$, it is easy to see $$ \frac{n}{n^2+k^2}\le\frac{n}{n^2-k^2x}\le\frac{n}{n^2-k^2a}. $$ Since $\sum_{k=1}^n\frac{n}{n^2+k^2}$ and $\sum_{k=1}^n\frac{n}{n^2-k^2a}$ converge to $\frac{\pi}{4}$ and $\int_0^1\frac{1}{1-t^2a}dt$ respectively, one has that $f_n(x)=\sum_{k=1}^n\frac{n}{n^2-k^2x}$ converges uniformly to $\int_0^1\frac{1}{1-t^2x}dt$ for $x\in[-1,a]$. Therefore $\sum_{k=1}^n\frac{n}{n^2-k^2x}$ converges to $\int_0^1\frac{1}{1-t^2x}dt$ for $x\in[-1,1)$ pointwise. Clearly $x=1$, $f_n(1)$ does not exist and hence $f_n(x)$ does not uniformly to $f(x)$ in $(-1,1)$.

Answer 2

Ok. I have resolved the problem at my standard. As pointed, $f_n$ do converges to the function $f:(-1,1) \longrightarrow \mathbb{R}$, $x \mapsto \int_0^1 \frac{1}{1+xt^2} dt$ since for each fixed $x \in (-1,1)$, the integrand $t \mapsto \frac{1}{1+xt^2}$ is continuous on $[0,1]$. By the elementary theorem which links the Riemann integration and the limit of the uniformly spaced Riemann sum, the result follows.

Now let us consider the difference between the difference between $f_n(x)$ and $f(x)$. Looking at the integrand $\frac{1}{1+xt^2}$, we can observe the convexity. It is concave above what $x<0$ and increasing. Hence the Riemann sum from the right endpoint exceeds the actual integration no less than $\left(\frac{1}{1+x}-1 \right) \times \frac{1}{2n}$. See the image below.

Therefore, for any given positive epsilon, as $x$ approaches to $-1$, we need more term than any given number to get close approximation to the actual integration correct at least the given epsilon. Hence the sequence does not converges uniformly.

I hope this to be a good exercise for undergraduate level students.