Limits of Sequences of functions and uniform convergence 3

Question

I have the sequence of functions: $s_n(x)= \displaystyle\sum_{k=0}^n \frac{k^2x}{1+k^4x^2}$ where $x \in [0,1]$

Consider$ [\delta, 1]$ for any $\delta \in (0,1]$

Now $0 \leq \frac{k^2x}{1+k^4x^2}\leq \frac{1}{k^2x} \leq \frac{1}{k^2\delta} $ So for any $x \in [\delta,1]$ The series converges uniformly on $[\delta,1]$ by the Weierstrass M-Test.

I know if $s_n$ is continuous the limit function $s(x)$ is also continuous on the same interval if there is uniform convergence . My problem is now the whole of $(0,1]$ The hint says show the limit function $s(x) = \displaystyle\sum_{k=0}^\infty \frac{k^2x}{1+k^4x^2}$ is not continuous on $[0,1]$. Why would I need to show it's not continuous on $[0,1]$? Surely it's on $(0,1]$?. My suspicion is that $s(x)$ looks a bit like the graph $\frac{1}{x}$ because we got the upper bound of $\frac{1}{x}\displaystyle \sum_{k=0}^\infty \frac{1}{k^2}$ and also the series diverges at $x=0$ so it would only be cont on (0,1] but not cont on $[0,1]$ since not defined at $0$ but this I'm not sure if this is right and its not really good enough. Could someone please help?

Answer 1

Hint: Consider the sequence $\{ 1/n^2 \}_{n=1}^\infty$. Can we get a lower bound bigger than $0$ on $s(1/n^2)$? If so, then $\lim_{x\to 0} s(x) \neq 0 = s(0)$.

The point to demonstrating that this is not continuous on $[0,1]$ is to show that $s_n(x)$ does not converge uniformly to $s(x)$, since otherwise $s(x)$ would be continuous on $[0,1]$.

Answer 2

Addressing the question "Why would I need to show it's not continuous on $[0,1]$? Surely it's on $(0,1]$?"

In fact, $s$ is continuous on $(0,1]$. To see this, choose $x \in (0,1]$ and $0 < \delta < x$. Since $(s_n)$ converges uniformly on $[\delta, 1]$, the limit function $s$ is continuous on $[\delta, 1]$, hence continuous at $x$.

Here is why it suffices to show that $s$ is not continuous on $[0,1]$, in order to conclude that $(s_n)$ is not uniformly convergent on $(0,1]$.

Let $\epsilon > 0$.

Suppose, for a contradiction, that $(s_n)$ converges uniformly on $(0,1]$. Then there is some $N$ such that $|s_n(x) - s(x)| < \epsilon$ for all $n > N$ and all $x \in (0,1]$.

Now fix any $n > N$. Since $s_n$ is continuous with $s_n(0) = 0$, there is a $\delta > 0$ such that $|s_n(x)| < \epsilon$ for all $0 \leq x < \delta$.

Combining the above, using the same fixed $n > N$, we have $|s(x)| \leq |s_n(x) - s(x)| + |s_n(x)| < 2\epsilon$ for all $0 < x < \delta$. This, along with the fact that $s(0) = 0$, shows that $s$ is continuous at $x=0$.

This means that if you can show that $s$ is not continuous at $x=0$, then our assumption that $(s_n)$ converges uniformly on $(0,1]$ must be false.