# For a branch-defined function, analyze the uniform convergence of it and of its series

Let $$(f_n)$$ be a sequence of functions on $$\mathbb{R}$$ such that $$f_n(x) = \left\{ \begin{array}{ll} 0, x \leq 0, \\ \frac{x}{n^2}, x \in (0,n^2), \\ 1, x \geq n^2. \end{array}\right.$$

a) Show that $$(f_n)$$ converges point-wise on $$\mathbb{R}$$.

b) Determine whether $$(f_n)$$ converges uniformly on $$\mathbb{R}$$.

c) Show that $$\displaystyle \sum_{n=1}^\infty f_n$$ converges point-wise on $$\mathbb{R}$$.

d) Show that $$\displaystyle \sum_{n=1}^\infty f_n$$ is uniformly convergent on every interval $$[0,a], a > 0$$, but it is not uniformly convergent on $$\mathbb{R}$$.

e) Show that $$s:\mathbb{R} \to \mathbb{R}$$ given by $$s(x) = \sum_{n=1}^\infty f_n(x)$$ is continuous.

For a), I think that we can easily see that $$f_n$$ tends to the function $$f \equiv 0$$ for all $$x \in \mathbb{R}$$ as $$n \to \infty$$.

For b), I am not sure how to determine whether $$(f_n)$$ is uniformly convergent. Do we just try and find out if $$\displaystyle \sup_{x \in \mathbb{R}} \{f_n(x) - 0 \} \to 0$$ as $$n \to \infty$$? Because if that's how it should be solved, then the function does not converge uniformly, since the supremum is $$1$$.

I think that for c) we have that for all $$x \in \mathbb{R}$$, the series $$\displaystyle \sum_{n =1}^\infty f_n(x)$$ contains a finite number of 1's and the other terms are $$\frac{x}{n^2},$$ so I think it's safe to assume that the series converges point-wise?

I do not know how to start d) (maybe for the interval $$[0,a]$$, we can apply the Weierstrass M-test). My intuition says that there is something bad happening when $$x \to \infty$$. The same goes for e).

• Hint for $b.)$ what is $f_n(n^2)-f(n^2)$? Nov 28, 2018 at 22:52
• That's exactly what I thought, I also edited the question right about now.
– user606835
Nov 28, 2018 at 22:53
• Uniform convergence doesn't mean for $\sum_{n=1}^\infty f_n(x)$. Possibly you mean $\sum_{k=1}^n f_k(x)$ right? Nov 29, 2018 at 8:43

(b) $$f_n$$ is not uniformly convergent. We show that this is not the case even when $$x\in (0,n^2)$$. Recall the definition of uniform convergence:$$\forall \epsilon>0\quad,\quad \exists N\quad,\quad\forall n>N\quad,\quad |{x\over n^2}|<\epsilon$$therefore $$n>{\sqrt{x\over \epsilon}}$$this means that we must choose $$N=\lfloor {\sqrt{x\over \epsilon}}\rfloor +1$$ which is surely dependent to $$x$$ not only to $$\epsilon$$, therefore $$f_n(x)$$ is not uniformly convergent.
(c) the statement is false since $$s(x)$$ is no function of $$n$$, but for $$s_n(x)$$ with the following definition$$s_n(x)=\sum_{k=1}^{n} f_k(x)$$we have $$s_n(x)\to s(x)$$where $$s(x)=\sum_{k=1}^{\infty} f_k(x)=\begin{cases}0&,\quad x<0\\\sum_{n=1\\ n>\sqrt x}^{\infty}{x\over n^2}+\sum_{n=1\\n\le \sqrt x}1&,\quad x\ge 0\end{cases}$$here is a sketch of $$s(x)$$ As you see, the function is continuous (which can be proved easily only in points $$x$$ with $$x=k^2$$ for some $$k\in \Bbb Z$$) and piecewise linear whose slope on each piece (from $$k^2$$ to $$(k+1)^2$$ increases unboundedly) therefore the function is uniformly continuous on each $$(0,a)$$ for $$a\in \Bbb R^+$$but not on $$\Bbb R$$. The proofs are easy and straight forward. Finally we have
d,e) $$s(x)$$ is continuous though not uniformly and $$s_n(x)$$ tends to $$s(x)$$ pointwise
I think e) is false. In fact, $$s(1+\frac 1 k) \to \sum_{n=2}^{\infty} \frac 1 {n^{2}}+1$$ whereas $$s(1) = \sum_{n=2}^{\infty} \frac 1 {n^{2}}$$.
Your argument for a) is correct. For c) use the fact that $$f_n(x) \leq \frac x {n^{2}}$$ for all $$x \geq 0$$. This also proves uniform convergence on $$[0,a]$$ by M-test. Second part of d) follows from b).