# A sequence of continuous functions which is pointwise convergent to zero and not uniformly convergent on any interval.

The exercise is to construct a sequence of continuous functions $f_n:\mathbb{R}\rightarrow \mathbb{R}, n\in \mathbb{N}$ , which is pointwise convergent to $f(x)=0 , x\in \mathbb{R}$ and not uniformly convergent on any interval $(a,b)$.

I posted this question once and I got the answer, but I had forgotten to mention that $f_n$ must be continuous. The example of sequence of discontinuous functions looks like this:

Let $\{r_n\}_{n\geq 1}$ be an enumeration of the rationals, and define $f_n(x)$ as follows: $$f_n(x)=\begin{cases} 1,& x\in \{r_k\}_{k\geq n}\\ 0,& else \end{cases}$$

The person, who posted this, made the remark:

If you want to make the $\{f_n\}$ continuous, you can modify this example by using bump functions supported on $(r_n-2^{-n},r_n+2^{-n})$ in place of the spikes to $1$ at $r_n$.

I'm not realy sure what he meant. I was thinking of defining $g_k$ as a broken line or maybe more smooth function going through $(r_k-2^{-k},0), (r_k,1), (r_k+2^{-k},0)$ and zero on $\mathbb{R}\setminus [r_k-2^{-k},r_k+2^{-k}]$ and then defining $f_n(x)=\sup\{g_k(x):k\ge n\}$. But I don't know whether it converges to zero and whether $f_n$ are even continuous.

On the other hand I have found this topic A sequence of continuous functions on $[0,1]$ which converge pointwise a.e. but does not converge uniformly on any interval and there is only an example of a sequence converging to zero almost everywhere. That's why I begin to doubt that there exists a sequence converging to zero everywhere (that would mean there is a mistake in my book because the exercise says "construct such a sequence").

• Are you familiar with convolution and mollifiers? – Silvia Ghinassi Feb 10 '16 at 16:41
• @SilviaGhinassi: What is your idea, even if the OP is not familiar with these notions? (I am interested in the answer, too). – PhoemueX Feb 10 '16 at 18:35
• @PhoemueX I didn't check if it really works, and I won't have access to a computer in the next few hours. I'll try to think more about it as soon as I can. Sorry about it. – Silvia Ghinassi Feb 10 '16 at 18:38
• The remark you quote looks interesting, but I don't see how that could work. If we add the "spikes" , then the sum diverges. – leonbloy Feb 11 '16 at 18:43
• @SilviaGhinassi with the current structure of rational numbers and irrational numbers the mollifier method can be hard to use, since the integral is zero. Perhaps we need a further discussion, not only to use the "rational-irrational" structure, but perhaps with some step functions. – Peter Feb 12 '16 at 17:37

According to the hint in "Problems in Mathematical Analysis", Biler and Witkowski, problem 4.112.

Define $g_n$ as the function which is $0$ on the intervals $(-\infty, 0]$ and $[2/n,\infty)$, $1$ at $1/n$, and linear otherwise. Then $g_n$ converges pointwise to $0$ but is not uniform in any interval containing $0$.

Now define $$f_n(x) = \sum_{k=0}^{\infty} 2^{-k} g_n(x - r_k).$$

Then the $f_n$ are continuous, converge pointwise to $0$, but do not converge uniformly on any open interval.

• Could you help me with the proof? I managed to prove that $f_n$ are well defined and continuous and $(f_n(x))$ converges to zero for all $x$. But I have problems with showing that this sequence doesn't converge uniformly on any interval. – Kulisty Feb 13 '16 at 22:31
• And just to make sure: Is $(r_n)$ an enumeration of the rationals? – Kulisty Feb 13 '16 at 23:10
• Yes, $r_n$ is an enumeration of the rationals. For nonuniform convergence, does it make sense that the convergence of $g_n$ is not uniform on any interval containing $0$? (For any interval containing $0$ and any number $N$, there is an $n > N$ and $x$ in the interval such that $g_n(x)=1$.) Similarly, for the nonuniform convergence of $f_n$: given any interval, there is a rational $r_k$ inside the interval, so for any number $N$ there is an $n > N$ and $x$ in the interval such that $f_n(x) \geq 2^{-k}$. – Michael Harrison Feb 14 '16 at 4:08
• Thank you very much, nice answer. I just came up with a different construction using Baire. I will award you the bounty if no other answers come up. – PhoemueX Feb 16 '16 at 9:38

There is indeed such a sequence, as one can show using a Baire category argument, using a suitable Banach space.

Let $C_{b}\left(\mathbb{R}\right)$ denote the space of bounded, continuous functions and let $$V:=\left\{ \left(f_{n}\right)_{n\in\mathbb{N}}\in\left(C_{b}\left(\mathbb{R}\right)\right)^{\mathbb{N}}\,\middle|\,\left\Vert f\right\Vert _{V}:=\sup_{n\in\mathbb{N}}\left\Vert f_{n}\right\Vert _{\infty}<\infty\right\} .$$ It is not hard to see that this is a Banach space, equipped with $\left\Vert \cdot\right\Vert _{V}$. Furthermore, define $$V_{0}:=\left\{ \left(f_{n}\right)_{n\in\mathbb{N}}\in V\,\middle|\,\forall x\in\mathbb{R}:\quad f_{n}\left(x\right)\xrightarrow[n\to\infty]{}0\right\} .$$ Then $V_{0}$ is complete, since it is a closed subspace of $V$. To see this, let $\left(g^{\left(m\right)}\right)_{m\in\mathbb{N}}=\left(\left(g_{n}^{\left(m\right)}\right)_{n\in\mathbb{N}}\right)_{m\in\mathbb{N}}$ be a sequence in $V_{0}$ with $g^{\left(m\right)}\xrightarrow[m\to\infty]{}g=\left(g_{n}\right)_{n\in\mathbb{N}}\in V$. We want to show $g\in V_{0}$. To this end, let $x\in\mathbb{R}$ and $\varepsilon>0$ be arbitrary. Then, there is some $m_{0}\in\mathbb{N}$ with $\left\Vert g^{\left(m\right)}-g\right\Vert _{V}<\frac{\varepsilon}{2}$ for all $m\geq m_{0}$. Because of $g^{\left(m_{0}\right)}\in V_{0}$, we have $g_{n}^{\left(m_{0}\right)}\left(x\right)\xrightarrow[n\to\infty]{}0$, so that there is $n_{0}\in\mathbb{N}$ with $\left|g_{n}^{\left(m_{0}\right)}\left(x\right)\right|<\frac{\varepsilon}{2}$ for all $n\geq n_{0}$. But for $n\geq n_{0}$, this implies \begin{align*} \left|g_{n}\left(x\right)\right| & \leq\left|g_{n}\left(x\right)-g_{n}^{\left(m_{0}\right)}\left(x\right)\right|+\left|g_{n}^{\left(m_{0}\right)}\left(x\right)\right|\\ & \leq\left\Vert g-g^{\left(m_{0}\right)}\right\Vert _{V}+\frac{\varepsilon}{2}\\ & <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Hence, $g_{n}\left(x\right)\xrightarrow[n\to\infty]{}0$. Since $x\in\mathbb{R}$ was arbitrary, we have shown $g\in V_{0}$, so that $V_{0}\leq V$ is closed and hence complete.

Now, let $\emptyset\neq I=\left(a,b\right)\subset\mathbb{R}$ be an arbitrary open, bounded, nonempty interval and set $$V_{0}^{\left(I\right)}:=\left\{ \left(f_{n}\right)_{n\in\mathbb{N}}\in V_{0}\,\middle|\,\left(f_{n}\right)_{n\in\mathbb{N}}\text{ does {\color{red}not} converge uniformly on }I\right\} .$$ I claim that $V_{0}^{\left(I\right)}\subset V_{0}$ is open and dense. Indeed,

1. $\left(V_{0}^{\left(I\right)}\right)^{c}\subset V_{0}$ is closed, because for a sequence $\left(g^{\left(m\right)}\right)_{m\in\mathbb{N}}=\left(\left(g_{n}^{\left(m\right)}\right)_{n\in\mathbb{N}}\right)_{m\in\mathbb{N}}$ in $\left(V_{0}^{\left(I\right)}\right)^{c}$ with $g^{\left(m\right)}\xrightarrow[m\to\infty]{}g=\left(g_{n}\right)_{n\in\mathbb{N}}\in V_{0}$, and $\varepsilon>0$, the following holds: There is some $m_{0}\in\mathbb{N}$ with $\left\Vert g-g^{\left(m\right)}\right\Vert _{V}<\frac{\varepsilon}{2}$ for all $m\geq m_{0}$. Furthermore, since $g^{\left(m_{0}\right)}\in\left(V_{0}^{\left(I\right)}\right)^{c}$, there is $n_{0}\in\mathbb{N}$ with $$\frac{\varepsilon}{2}>\left\Vert g_{n}^{\left(m_{0}\right)}\right\Vert _{\sup,I}:=\sup_{x\in I}\left|g_{n}^{\left(m_{0}\right)}\left(x\right)\right|$$ for all $n\geq n_{0}$. But this implies \begin{align*} \left\Vert g_{n}\right\Vert _{\sup,I} & \leq\left\Vert g_{n}-g_{n}^{\left(m_{0}\right)}\right\Vert _{\sup,I}+\left\Vert g_{n}^{\left(m_{0}\right)}\right\Vert _{\sup,I}\\ & \leq\left\Vert g-g^{\left(m_{0}\right)}\right\Vert _{V}+\left\Vert g_{n}^{\left(m_{0}\right)}\right\Vert _{\sup,I}\\ & <\varepsilon \end{align*} for all $n\geq n_{0}$ and thus $g=\left(g_{n}\right)_{n\in\mathbb{N}}\in\left(V_{0}^{\left(I\right)}\right)^{c}$.

2. To see that $V_{0}^{\left(I\right)}\subset V_{0}$ is dense, let $g=\left(g_{n}\right)_{n\in\mathbb{N}}\in V_{0}$ be arbitrary. We want to show $g\in\overline{V_{0}^{\left(I\right)}}$. If $g\in V_{0}^{\left(I\right)}$, this is clear. Hence, we can assume $g\notin V_{0}^{\left(I\right)}$, i.e. $g_{n}\to0$ uniformly on $I$. Now, let $\varepsilon>0$ be arbitrary. We want to find some $h=\left(h_{n}\right)_{n\in\mathbb{N}}\in V_{0}^{\left(I\right)}$with $\left\Vert g-h\right\Vert _{V}\leq\varepsilon$. To this end, let $f_{n}\equiv0$ for $n\leq\frac{2}{b-a}$, i.e. for $a+\frac{2}{n}\geq b$. For $n>\frac{2}{b-a}$, define $$f_{n}\left(x\right):=\begin{cases} 0, & \text{if }x\leq a,\\ \varepsilon\cdot n\left(x-a\right), & \text{if }a\leq x\leq a+\frac{1}{n},\\ \varepsilon\cdot\left(1-n\cdot\left(x-a-\frac{1}{n}\right)\right), & \text{if }a+\frac{1}{n}\leq x\leq a+\frac{2}{n},\\ 0, & \text{if }x\geq a+\frac{2}{n}. \end{cases}$$ It is clear that $f:=\left(f_{n}\right)_{n\in\mathbb{N}}\in V_{0}$ (pointwise convergence is clear for $x\leq a$. But for $x>a$, we have $x\geq a+\frac{1}{n}$ for all $n\geq n_{x}$ and hence $f_{n}\left(x\right)=0$) with $\left\Vert f\right\Vert _{V}=\varepsilon$. Furthermore, we have $\left\Vert f_{n}\right\Vert _{\sup,I}=\varepsilon$ for all $n>\frac{2}{b-a}$ and hence $\left(f_{n}\right)_{n\in\mathbb{N}}$ does not converge uniformly (to $0$) on $I$. Thus, neither does $h:=\left(h_{n}\right)_{n\in\mathbb{N}}:=\left(f_{n}+g_{n}\right)_{n\in\mathbb{N}}$, because otherwise, $f_{n}=h_{n}+\left(-g_{n}\right)\to0$ uniformly on $I$, by uniform convergence of $\left(g_{n}\right)_{n\in\mathbb{N}}$.

Altogether, $h\in V_{0}^{\left(I\right)}$ with $\left\Vert g-h\right\Vert _{V}=\left\Vert f\right\Vert _{V}=\varepsilon$, so that $V_{0}^{\left(I\right)}\subset V_{0}$ is dense.

Finally, let $$J:=\left\{ \left(a,b\right)\,\middle|\, a,b\in\mathbb{Q}\text{ and }a<b\right\} .$$ Then $J$ is countable, so that Baire's theorem shows that $$M:=\bigcap_{I\in J}V_{0}^{\left(I\right)}\subset V_{0}$$ is dense. In particular, $M\neq\emptyset$. But any $f=\left(f_{n}\right)_{n\in\mathbb{N}}\in M$ satisfies $f_{n}\to0$ pointwise, but $\left(f_{n}\right)_{n\in\mathbb{N}}$ does not converge uniformly (to $0$) on any interval from $J$. Since every (nontrivial) interval contains an interval from $J$, we are done.