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).

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

2 Answers 2


(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 $s(x)$ enter image description here 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).


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