# Thread: Cts functions on [0,1] tending ptwise to 0 - convergence uniform on an interval?

1. ## Cts functions on [0,1] tending ptwise to 0 - convergence uniform on an interval?

Hello everyone, first post! I've come across the following Analysis problem and given that it's one of the hardest questions on my problem sheet, I am having a few problems working my argument through.

"Suppose that f_n:[0,1] -> Reals is a sequence of continuous functions tending pointwise to 0. Must there be an interval on which f_n -> 0 uniformly?"

I have considered using the Weierstrass approximation theorem here, which states that we can find, for any continuous function [0,1] -> Reals, a uniform approximation by polynomials.

Because of this, it seems to me - though I could be wrong - that these f_n -> 0 uniformly if this series of polynomials (each p_n approximating an f_n to a sufficient degree of accuracy) tends to 0 uniformly - in which case it suffices to prove the result for any series of polynomials.

Even if this deduction -is- correct, which I'm not 100% confident about, I can't seem to follow through and show that there exists such an interval for a polynomial sequence. On the other hand, perhaps there is a counterexample and I'm going about this completely the wrong way! Could anyone lend a hand please?

2. I can see how the statement would be true if the family of functions was equicontinuous.

First, there is a point $\displaystyle x_0\in [0,1]$ such that $\displaystyle \{f_n(x_0)\}\to 0$. In other words, for any $\displaystyle \varepsilon>0$, there exists $\displaystyle N$ such that $\displaystyle n>N$ implies $\displaystyle |f_n(x_0)|<\varepsilon$. Since the $\displaystyle f_n$ are equicontinuous, there is $\displaystyle \delta>0$ such that $\displaystyle |x-x_0|<\delta$ implies $\displaystyle |f_n(x)-f_n(x_0)|<\varepsilon$ for all $\displaystyle f_n$. This gives the inequality

$\displaystyle |f_n(x)|=|f_n(x)-f_n(x_0)+f_n(x_0)|\leq |f_n(x)-f_n(x_0)|+|f_n(x_0)|<\varepsilon + \varepsilon=2\varepsilon$

whenever $\displaystyle x\in (x_0-\delta, x_0+\delta)$ and $\displaystyle n>N$. This proves that $\displaystyle \{f_n\}$ is uniformly convergent to 0 on the interval $\displaystyle (x_0-\delta, x_0+\delta)$.

At the moment, I am not sure if equicontinuity can be dropped.

3. Originally Posted by mathmos8128
Suppose that f_n:[0,1] -> Reals is a sequence of continuous functions tending pointwise to 0. Must there be an interval on which f_n -> 0 uniformly?
The answer is No, but I found it quite tricky to construct a counterexample.

Start with something easier. Define a sequence of functions $\displaystyle f_n(x)$ on the unit interval as follows: on the interval from 0 to $\displaystyle 2^{-n-1}$, $\displaystyle f_n(x)$ increases linearly from 0 to 1; on the interval from $\displaystyle 2^{-n-1}$ to $\displaystyle 2^{-n}$, $\displaystyle f_n(x)$ decreases linearly from 1 to 0; and on the remainder of the interval, $\displaystyle f_n(x) = 0$. Then it is well known that $\displaystyle f_n(x)\to0$ pointwise but not uniformly on the unit interval. In fact, if the binary expansion of x has a 1 in any of the first k places, then $\displaystyle f_n(x) = 0$ whenever $\displaystyle n\geqslant k$.

Now extend $\displaystyle f_n(x)$ by periodicity to a function on the whole real line, with period 1. Thus $\displaystyle f_n(x)$ is a function with a narrow spike, of height 1, to the right of each integer point.

Next, define a sequence of functions $\displaystyle g_n(x)$ by $\displaystyle g_n(x) = \sum_{k=1}^n 2^{-k}f_n(2^kx)$. Then $\displaystyle g_n(x)$ is a continuous function. If $\displaystyle n\geqslant k$ then $\displaystyle g_n(x) \geqslant 2^{-k}f_n(2^kx)$. But $\displaystyle 2^{-k}f_n(2^kx)$ attains the value $\displaystyle 2^{-k}$ in every interval of length $\displaystyle 2^{-k}$. Therefore $\displaystyle \max\{g_n(x):x\in J\}\geqslant 2^{-k}$ for all $\displaystyle n\geqslant k$ in any interval $\displaystyle J$ of length $\displaystyle 2^{-k}$. Therefore $\displaystyle g_n(x)$ does not tend uniformly to 0 in any interval.

However, $\displaystyle g_n(x)\to0$ pointwise. To see that, we have to consider two separate cases.

Case 1: x has only finitely many 1's in its binary expansion. Suppose that the last 1 occurs in the r'th place in the binary expansion, and let $\displaystyle n>r$. Then for $\displaystyle k<r$, $\displaystyle 2^kx$ has a 1 somewhere in the first n places of its binary expansion, and so $\displaystyle f_n(2^kx)=0$. And for $\displaystyle k\geqslant r$, $\displaystyle 2^kx$ is an integer, so again $\displaystyle f_n(2^kx)=0$. Thus $\displaystyle g_n(x) = 0$ for all $\displaystyle n> r$.

Case 2: x has infinitely many 1's in its binary expansion. Then given $\displaystyle \varepsilon>0$, there exists $\displaystyle r$ with $\displaystyle 2^{-r} < \varepsilon/2$ such that x has a 1 in its $\displaystyle r$'th binary place. If $\displaystyle n>r$ then $\displaystyle g_n(x) = \sum_{k=1}^r 2^{-k}f_n(2^kx) + \sum_{k=r+1}^n 2^{-k}f_n(2^kx)$. But the first of those sums is 0, and the second is less than $\displaystyle \varepsilon$. Therefore $\displaystyle g_n(x) < \varepsilon$.

So in either of the two cases, $\displaystyle g_n(x)\to0$ as $\displaystyle n\to\infty$.

4. Originally Posted by Opalg
The answer is No, but I found it quite tricky to construct a counterexample.

Start with something easier. Define a sequence of functions $\displaystyle f_n(x)$ on the unit interval as follows: on the interval from 0 to $\displaystyle 2^{-n-1}$, $\displaystyle f_n(x)$ increases linearly from 0 to 1; on the interval from $\displaystyle 2^{-n-1}$ to $\displaystyle 2^{-n}$, $\displaystyle f_n(x)$ decreases linearly from 1 to 0; and on the remainder of the interval, $\displaystyle f_n(x) = 0$. Then it is well known that $\displaystyle f_n(x)\to0$ pointwise but not uniformly on the unit interval. In fact, if the binary expansion of x has a 1 in any of the first k places, then $\displaystyle f_n(x) = 0$ whenever $\displaystyle n\geqslant k$.

Now extend $\displaystyle f_n(x)$ by periodicity to a function on the whole real line, with period 1. Thus $\displaystyle f_n(x)$ is a function with a narrow spike, of height 1, to the right of each integer point.

Next, define a sequence of functions $\displaystyle g_n(x)$ by $\displaystyle g_n(x) = \sum_{k=1}^n 2^{-k}f_n(2^kx)$. Then $\displaystyle g_n(x)$ is a continuous function. If $\displaystyle n\geqslant k$ then $\displaystyle g_n(x) \geqslant 2^{-k}f_n(2^kx)$. But $\displaystyle 2^{-k}f_n(2^kx)$ attains the value $\displaystyle 2^{-k}$ in every interval of length $\displaystyle 2^{-k}$. Therefore $\displaystyle \max\{g_n(x):x\in J\}\geqslant 2^{-k}$ for all $\displaystyle n\geqslant k$ in any interval $\displaystyle J$ of length $\displaystyle 2^{-k}$. Therefore $\displaystyle g_n(x)$ does not tend uniformly to 0 in any interval.

However, $\displaystyle g_n(x)\to0$ pointwise. To see that, we have to consider two separate cases.

Case 1: x has only finitely many 1's in its binary expansion. Suppose that the last 1 occurs in the r'th place in the binary expansion, and let $\displaystyle n>r$. Then for $\displaystyle k<r$, $\displaystyle 2^kx$ has a 1 somewhere in the first n places of its binary expansion, and so $\displaystyle f_n(2^kx)=0$. And for $\displaystyle k\geqslant r$, $\displaystyle 2^kx$ is an integer, so again $\displaystyle f_n(2^kx)=0$. Thus $\displaystyle g_n(x) = 0$ for all $\displaystyle n> r$.

Case 2: x has infinitely many 1's in its binary expansion. Then given $\displaystyle \varepsilon>0$, there exists $\displaystyle r$ with $\displaystyle 2^{-r} < \varepsilon/2$ such that x has a 1 in its $\displaystyle r$'th binary place. If $\displaystyle n>r$ then $\displaystyle g_n(x) = \sum_{k=1}^r 2^{-k}f_n(2^kx) + \sum_{k=r+1}^n 2^{-k}f_n(2^kx)$. But the first of those sums is 0, and the second is less than $\displaystyle \varepsilon$. Therefore $\displaystyle g_n(x) < \varepsilon$.

So in either of the two cases, $\displaystyle g_n(x)\to0$ as $\displaystyle n\to\infty$.
Wow, that's genius! If you don't mind me asking, how did you come up with that counterexample? It's certainly not something I could just construct off the top of my head! Thankyou so much

5. Originally Posted by mathmos8128
If you don't mind me asking, how did you come up with that counterexample?
The starting point was the Wikipedia article on Dini's theorem, which makes it pretty clear that pointwise convergence only implies uniform convergence if you have a strong additional condition, such as the sequence being monotonic. That made me confident enough to look for a counterexample rather than trying to prove a positive result. Finding a counterexample that actually worked was something that I had to think about overnight. The obvious strategy is to take a sequence that converges pointwise but not uniformly (usually because something blows up at a single point), and then find some way of spreading that pathology throughout an interval. But I had several false starts before getting the details right.

6. Originally Posted by Opalg
The starting point was the Wikipedia article on Dini's theorem, which makes it pretty clear that pointwise convergence only implies uniform convergence if you have a strong additional condition, such as the sequence being monotonic. That made me confident enough to look for a counterexample rather than trying to prove a positive result. Finding a counterexample that actually worked was something that I had to think about overnight. The obvious strategy is to take a sequence that converges pointwise but not uniformly (usually because something blows up at a single point), and then find some way of spreading that pathology throughout an interval. But I had several false starts before getting the details right.
Well that is extremely clever, I'd never heard of Dini's theorem before - I had actually looked at the idea of 'moving triangles' but I couldn't find a way to put them together to provide a solid counterexample - thankyou very much for all the effort! It's geniunely appreciated.