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Let $f:\mathbb R \to \mathbb R$ be a continuous function. Define $f_n:\mathbb R \to \mathbb R$ by $$ f_n(x) := f(x+1/n). $$ Suppose that $(f_n)_{n=1}^\infty$ converges uniformly to $f$. Does it follow that $f$ is uniformly continuous?

Note: the answer is clearly no if we don't assume that $f$ is continuous. I suspect there is a counterexample, showing that the answer is no even if $f$ is continuous.

Edit: The following observation might help.

There exists a continuous, non-uniformly continuous function $f:\mathbb R \to \mathbb R$ such that $(f_n)_{n=1}^\infty$ converges uniformly to $f$, if and only if there exists a continuous, non-uniformly continuous function $g:\mathbb N \times \mathbb R \to \mathbb R$, such that $(g_n)_{n=1}^\infty$ converges uniformly to $g$, where $g_n(k,x) = g(k,x+1/n)$ and the metric on $\mathbb N \times \mathbb R$ comes from viewing it as a subspace of $\mathbb R \times \mathbb R$ (with the Euclidean metric, say).

Proof: Given the function $g$, there is some $\epsilon>0$ which witnesses non-uniform continuity. That is, for each $n$, there exists $m \in \mathbb N$ and $x,y \in \mathbb R$ such that $|x-y|<1/n$ and $$|g(m,x)-g(m,y)|\geq\epsilon.$$ By moving things around, we can assume that for each $k$, there exists $y \in (0,1/k)$ such that $$ |g(k,0)-g(k,y)| \geq \epsilon. $$

Next, we can also assume that $g(k,x)=0$ for all $|x|>2$. This is achieved by multiplying $g$ by the function $h$ which is $1$ on $\mathbb N \times [-1,1]$, $0$ on $\mathbb N \times (\mathbb R \setminus (-2,2)$, and linear elsewhere. That $(gh)_n \to gh$ uniformly can be proven by the same argument that the product of uniformly continuous functions is uniformly continuous.

Now, we just define $f$ piecewise, by [ f(6k+x)=g(k,x) ] for $k \in \mathbb N$ and $x \in [-3,3]$, and [ f(x) = 0 ] for $x < -3$.

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    $\begingroup$ Hello and welcome to math.stackexchange. That's a very nice question! $\endgroup$ – Hans Engler Feb 25 '16 at 18:34
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    $\begingroup$ Clement, that is a good idea. I can't see how to prove it even in the case that $f$ is continuously differentiable. There is a Mean Value Theorem argument if $f$ has a uniformly continuous derivative, but it breaks down outside of that case. $\endgroup$ – Aaron T Feb 26 '16 at 20:57
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    $\begingroup$ Have you thought about your question when $f_{n}(x) = f(x + 2^{-n})$? $\endgroup$ – fourierwho Jul 30 '16 at 19:29
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    $\begingroup$ My first reaction was that this must be trivial. But I don't see exactly how to do it... $\endgroup$ – David C. Ullrich Aug 3 '16 at 15:08
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    $\begingroup$ Actually, Milo's proof can be extended to prove this. $\endgroup$ – George Lowther Aug 4 '16 at 23:44
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The given conditions do not imply uniform continuity.

Let us construct our counterexample by first creating a "weight" function on the rational numbers defined as follows: $$w(x)=\min\left\{n:x=a_0+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n},\,a_i\in\mathbb Z\right\}$$ That is, $w(x)$ is the least number $n$ such that $x$ is an integer plus the reciprocal of $n$ integers. Note that $w$ is $0$ on the integers themselves. Now, define a family of functions $d_{\alpha,\beta}:\mathbb R\rightarrow\mathbb R$ for $\alpha,\beta>0$ as follows: $$d_{\alpha,\beta}(x)=\min\{\alpha|x-q|+\beta w(q):q\in \mathbb Q\}.$$ One can see that the given set actually has a minimum since the set $Q_n=\{q\in \mathbb Q:w(q)\leq n\}$ is closed, so there is a well-defined minimum distance from $x$ to an element of $Q_n$, and for large $n$ the $\beta w(q)$ term will be too large to be a minimum. We may also note that this function satisfies the triangle inequality: $$d_{\alpha,\beta}(x+y)\leq d_{\alpha,\beta}(x)+d_{\alpha,\beta}(y).$$ This may be proved by first noting that $w$ satisfies the triangle inequality, then doing some algebra.

We will critically rely on two consequences of the triangle inequality. Firstly, since $d_{\alpha,\beta}(x)\leq \alpha |x|$, we get that $d_{\alpha,\beta}$ is continuous. We can also use the fact that $d_{\alpha,\beta}(1/n)=\min(\beta,\alpha/n)$ as a bound for the difference $d_{\alpha,\beta}(x)-d_{\alpha,\beta}(x+1/n)$, which is important for our application.

Now, let us choose a sequence $s_i$ of irrational numbers converging to $0$ and try to set our parameters to make $d_{\alpha,\beta}(s_m)$ large, while decreasing $\beta$. Thus, we get that differences $d_{\alpha,\beta}(x)-d_{\alpha,\beta}(x+1/n)$ will be small, but the function will obtain large values near zero. The big picture is that we will stitch together lots of functions like this, causing a big problem for uniform continuity.

Specifically, let $\varepsilon_i$ be the distance from $s_i$ to any element of $Q_i$. That is, $\varepsilon_i=\min\{|s_i-q|:q\in Q_i\}$. Set $\alpha_i=\frac{1}{\varepsilon_i}$ and $\beta_i=\frac{1}i$. We may note that $d_{\alpha_i,\beta_i}(s_i)\geq 1$ by seeing that if $q\in Q_i$, then $\alpha_i|s_i-q|\geq 1$ by definition of $\alpha_i$, and if $q\not\in Q_i$ then $\beta_i w(q)\geq 1$, thus the expression $\alpha_i|s_i-q|+\beta_i w(q)$ is at least $1$ for all rational $q$.

Now, we need to come up with a way to get our function to be compactly supported, but still do what we want. Define a window map $g$ by the following relations: $$g(x)=\begin{cases}x & \text{if }0\leq x\leq 1 \\ 1 & \text{if } 1\leq x\leq 2 \\ 3-x & \text{if }2\leq x \leq 3 \\ 0 &\text{otherwise}\end{cases}.$$ This map is a sort of "table" with a flat middle section and two sloping sides. Now, we define $$f_i(x)=g(x)\cdot \min(d_{\alpha_i,\beta_i}(x),1).$$ Let us bound $|f_i(x)-f_i(x+1/n)|$. To do so, note that $\left| d_{\alpha_i,\beta_i}(x)-d_{\alpha_i,\beta_i}(x+1/n)\right|\leq \min(\beta_i,\alpha_i/n)$. Obviously, $\left|g(x)-g(x+1/n)\right|\leq 1/n$. Together, we can use these to get the bound: $$\left|f_i(x)-f_i(x+1/n)\right|\leq 1/n + \min(\beta_i,\alpha_i/n).$$ We also have that $|f_i(1)-f_i(1+s_i)|=1$, which we will use later to contradict uniform continuity.

Finally, we do the stitching. Define $f$ as follows: $$f(x)=\begin{cases}f_i(x-3i) & \text{if }3i \leq x \leq 3i+3,\,i\in \mathbb N \\ 0 &\text{if } x \leq 0 \end{cases}.$$ Since, in any interval of length at most one, $f$ may decomposed as a sum of translates of two functions $f_i$ and $f_{i+1}$, we immediately get an inequality $$\left|f(x)-f(x+1/n)\right| \leq 2/n + \min(\beta_i,\alpha_i/n) + \min(\beta_{i+1},\alpha_{i+1}/n).$$ $$\left|f(x)-f(x+1/n)\right| \leq \sup \{2/n + 2\min(\beta_i,\alpha_i/n):i\in\mathbb N\}$$ One has that $\lim_{n\rightarrow\infty}\sup\{2/n + 2\min(\beta_i,\alpha_i/n)\}=0$ since, for any $\varepsilon$, there is some $I$ such that for all $i>I$ we have $\beta_i<\varepsilon$. However, if we let $N=\max\{\alpha_i/\varepsilon:i\leq I\}$ then we have that $\alpha_i/n<\varepsilon$ for all $i\leq I$ and $n>N$. Thus, the given term is less than $2/n+2\varepsilon$ for all $n>N$, meaning it converges to $0$. This, of course, gives that $f(x+1/n)$ converges uniformly to $f(x)$.

However, $f$ is not uniformly continuous since $|f(3i+1)-f(3i+1+s_i)|=1$ for all $s_i$, even though $s_i$ goes to zero. Thus, there can be no $\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<1$. We are therefore done.


I thought, to help guide intuition, a few pictures might help out. One should note that only the ratio $\alpha/\beta$ actually affects the shape of $d_{\alpha,\beta}$ due to the identity $d_{c\alpha,c\beta}(x)=cd_{\alpha,\beta}(x)$. Here's a couple plots where $\alpha/\beta=12$ in the first picture, $40$ in the second, and $600$ in the third. Each one shows an increase in the number of "tiers" in the function due to bands of rationals where $w(q)=d_{\alpha,\beta}(q)$. enter image description here enter image description here enter image description here The last picture nearly melted my computer to plot. The moral of the story there is that $\alpha_i$ might need to be really big to satisfy what is demanded of it.

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  • $\begingroup$ Looks good to me! $\endgroup$ – George Lowther Aug 4 '16 at 23:45
  • $\begingroup$ I think some of the of the i's and n's are mixed up though $\endgroup$ – George Lowther Aug 4 '16 at 23:49
  • $\begingroup$ @GeorgeLowther I think I fixed the mixed up indices. $\endgroup$ – Milo Brandt Aug 5 '16 at 0:16
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    $\begingroup$ @Nobody Well, the intuition I started with is not too bad: Uniform convergence means that the function can't change it's value too much due to a translation by $1/n$. Thus, it makes sense to measure the "distance" between points by how many such translations it takes. That's $w(q)$ here, and we want to think of the metric $d(x,y)=w(x-y)$. So I wanted to create a function that played nice in that metric, but was also continuous - and, after a few hours of pondering, I thought of just demanding $d(x,y)\leq \alpha |x-y|$ to give $d_{\alpha,\beta}$. The rest was just bookkeeping. $\endgroup$ – Milo Brandt Aug 5 '16 at 3:25
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    $\begingroup$ Nice answer, thank you. $\endgroup$ – Aaron T Aug 5 '16 at 13:08
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I have very partial results, but maybe somebody will be able to use some my ideas to obtain the bounty. :-) I was thinking about this problem and I suggest that, in general, the answer is negative, but, maybe a counterexample is a fractal function, as the Koch curve or Cantor staircase. The rest of the answer is devoted for positive results for restricted cases.

For each $n$ put $$\varepsilon(n)=\|f-f_n\|=\sup_{x\in\Bbb R} |f(x+1/n)-f(x)|.$$ We are given that the sequence $\{\varepsilon(n)\}$ of non-negative numbers tends to zero when $n$ tends to the infinity.

A cheap remark. If the function $f$ is monotonic then the answer is positive. Indeed, let $\varepsilon>0$ be an arbitrary number. Choose a number $n$ such that $\varepsilon(n)<\varepsilon$. Let $x$ and $x’$ be arbitrary reals such that $x<x’<x+1/n$.

Then by monotonicity of the function $f$ we have $$|f(x’)-f(x)|< |f(x+1/n)-f(x)|\le\varepsilon_n<\varepsilon.$$

The remaining part of the answer is less trivial.

Assume that there exists a natural number $k\ge 2$ such that a series $\sum_{i=1}^\infty\varepsilon(k^i)$ converges. I claim that then the answer is positive. Indeed, let $\varepsilon>0$ be an arbitrary number. Choose a number $N\ge 1$ such that $\sum_{i=N}^\infty\varepsilon(k^i)<\varepsilon/k$. Put $\delta=k^{-N}$. Let $x$ and $x’$ be arbitrary reals such that $x<x’<x+\delta$. Since the function $f$ is continuous at the point $x’$, there exists a number $0<\delta’<x’-x$ such that if $|x’’-x’|<\delta’$ then $|f(x’’)-f(x’)|<\varepsilon$. Pick any number $x’’$ such that $x’-\delta’<x’’<x’$ and $k^{p}(x’’-x)$ is integer for some natural $p$, so the representation of the number $x’’-x$ in $k$-adic system is finite. Since each “digit” in this representation is at most $k-1$, we see that there exists finite subsets $S_1,\dots, S_k$ of the set $\{N,N+1,\dots, p\}$ such that $x’’-x=\sum_{j=1}^{k-1} \Delta_j$, where $\Delta_j=\sum_{i\in S_j} k^{-i}$. Then

$$|f(x’)-f(x)|\le |f(x’)-f(x’’)|+ |f(x’’)-f(x)|<\varepsilon+ \sum_{j=1}^{k-1}\sum_{i\in S_j} \varepsilon(k^i)\le \varepsilon+ \sum_{j=1}^{k-1}\varepsilon/k<2\varepsilon.$$

Similarly we can show a more general fact that the answer is positive provided there exists a natural number $M$ and an (increasing) sequence $\{k_i\}$ of natural numbers such that a series $\sum_{i=1}^\infty\varepsilon(k_i)$ converges and for each natural $N$ the set

$$\left\{\sum_{i=N}^\infty m_i/k_i: m_i\in\Bbb Z\cap [-M,M]\right\}$$

is dense in some neighborhood of zero.

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