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Suppose $f:\mathbb{R} \to \mathbb{R}$ is continuous and $\lim_{x \to \infty} [f(x+y) – f(x)] = 0$ pointwise for all $y \in \mathbb{R}$.

Can we show convergence is uniform for $y \in \mathbb{R}$? Can we show it is uniform on any compact set?

For the first one, I came up with a counterexample $f(x) = \sqrt{x}$ for $x \geq 0$ and $f(x) = 0$ for $x < 0$. Now for $x>0$ and $y > 0$, $f(x+y) – f(x) = \sqrt{x+y} - \sqrt{x} = \frac{y}{\sqrt{x+y} + \sqrt{x}}$ . The convergence is not uniform for all $y \in \mathbb{R}$ because for any $\epsilon $ and for large $x$ we can choose $y = x$ so that $f(x+y) – f(x) = \frac{x}{\sqrt{x+x} + \sqrt{x}} = \frac{\sqrt{x}}{\sqrt{2} + 1} > \epsilon$.

But how to prove the convergence is uniform for any continuous function $f$ on any compact set?

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    $\begingroup$ To begin with, $x$ and $y$ are both $\geq 0.$ If $y$ is bounded by a constant $M$ then $|f(x + y) - f(x)| \leq \dfrac{M}{2 \sqrt{x}}.$ Q.E.D. $\endgroup$
    – William M.
    Commented Mar 1, 2019 at 22:32
  • $\begingroup$ @WillM: I don't think it is that simple. I understand what you showed for $f(x) = \sqrt{x}$ on a compact set. But that function is just a specific counterexample for uniform convergence on $\mathbb{R}$. The function in the question is an arbitrary continuous function. $\endgroup$
    – scobaco
    Commented Mar 3, 2019 at 21:53
  • $\begingroup$ Oh, I see. I got confused with his wording. $\endgroup$
    – William M.
    Commented Mar 3, 2019 at 23:23

2 Answers 2

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Claim: Let $A=[0,r]$ with some fixed $r>0$ and suppose that $\lim\limits_{x\to\infty}(f(x+y)-f(x))=0$ for every $y\in A$. Then $\lim\limits_{x\to\infty}(f(x+y)-f(x))=0$ uniformly for $y\in A$.

For $\varepsilon>0$ and $n\in\mathbb{N}$, define $A_n(\varepsilon)=\Big\{y\in A: \exists x\ge n \quad \big|f(x+y)-f(x)\big|>\varepsilon\Big\}$. (This set is open.) An equivalent form of the Claim is that for every $\varepsilon>0$ there is some $n\in\mathbb{N}$ such that $A_n(\varepsilon)=\emptyset$.


Lemma: If $A_n(\varepsilon)\ne\emptyset$ then $\lambda\big(A_n(\frac\varepsilon2)\big)\ge \frac{r}{4}$.

Proof for the Lemma: Take a $y_0\in A_n(\varepsilon)$ and some $x_0\ge n$ such that $\big|f(x_0+y_0)-f(x_0)\big|>\varepsilon$.

Case 1: The $y_0$ is "big", $y_0\ge\frac{r}{2}$.

For every $0<y<y_0$, by the triangle inequality we get $$ \big|f(x_0+y)-f(x_0)\big|+\big|f(x_0+y_0)-f(x_0+y)\big| \ge \big|f(x_0+y_0)-f(x_0)\big|>\varepsilon $$ $$ \big|f(x_0+y)-f(x_0)\big|>\frac\varepsilon2 \quad\text{or}\quad \big|f(x_0+y_0)-f(x_0+y)\big|>\frac\varepsilon2 $$ $$ y\in A_n(\tfrac\varepsilon2) \quad\text{or}\quad y_0-y\in A_n(\tfrac\varepsilon2) $$ $$ y\in A_n(\tfrac\varepsilon2) \quad\text{or}\quad y\in \Big(y_0-A_n(\tfrac\varepsilon2)\Big). $$ Hence, the set $A_n(\frac\varepsilon2)$ and its reflection $y_0-A_n(\frac\varepsilon2)$ cover the interval $(0,y_0)$, and therefore $\lambda\big(A_n(\frac\varepsilon2)\big)\ge\frac12\lambda((0,y_0)) =\frac{y_0}{2}\ge\frac{r}{4}$.

Case 2: The $y_0$ is "small", $0\le y_0<\frac{r}{2}$. Similarly to the first case, for every $y\in(0,r-y_0)$, $$ \big|f(x_0+y+y_0)-f(x_0+y_0)\big|+\big|f(x_0+y+y_0)-f(x_0)\big| \ge \big|f(x_0+y_0)-f(x_0)\big|>\varepsilon $$ $$ \big|f(x_0+y+y_0)-f(x_0+y_0)\big|>\frac\varepsilon2 \quad\text{or}\quad \big|f(x_0+y+y_0)-f(x_0)\big|>\frac\varepsilon2 $$ $$ y\in A_n(\tfrac\varepsilon2) \quad\text{or}\quad y+y_0\in A_n(\tfrac\varepsilon2) $$ $$ y\in A_n(\tfrac\varepsilon2) \quad\text{or}\quad y\in \Big(A_n(\tfrac\varepsilon2)-y_0\Big). $$ The sets $A_n(\frac\varepsilon2)$ and its translation $A_n(\frac\varepsilon2)-y_0$ cover $(0,r-y_0)$, and therefore $\lambda\big(A_n(\frac\varepsilon2)\big)\ge\frac12\lambda((0,r-y_0)) =\frac{r-y_0}{2}\ge\frac{r}{4}$.


Prove for the Claim: Fix an arbitrary $\varepsilon>0$. Since $f(x+y)-f(x)\to0$ pointwise, for every $y\in A$ there is some index $n$ such that $y\notin A_n(\tfrac\varepsilon2)$. So $\bigcap_n A_n(\tfrac\varepsilon2)=\emptyset$.

The sequence $A\supset A_1(\tfrac\varepsilon2)\supset A_2(\tfrac\varepsilon2)\supset\ldots$ is nonincreasing and the measures of the sets are finite, so $\lambda(A_n(\tfrac\varepsilon2))\to0$. Hence, there is an index $n$ with $\lambda(A_n(\tfrac\varepsilon2))<\frac{r}{4}$ and therefore $A_n(\varepsilon)=\emptyset$.

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  • $\begingroup$ Thank you. I'm working through the details, but this looks good so far. I was never comfortable with the first answer. There is nothing here to enforce equicontinuity of $f(x+y)$ for $y\in K$ compact and $x \in \mathbb{R}$. Am I correct? $\endgroup$
    – scobaco
    Commented Jan 8, 2020 at 17:40
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Maybe this result here comes in handy Equicontinuity on a compact metric space turns pointwise to uniform convergence Equicontinuity is easy to prove. Choose your sequence of functions to be $f_n(x)=f(x_n+x)$ where $x_n$ is some fixed sequence tending to $\infty$ and prove your statement by contradiction using the result in my link

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  • $\begingroup$ Look, maybe I am confused, maybe you are. You were asking something about compact sets, have you changed your goal now? $\endgroup$
    – Sorin Tirc
    Commented Mar 3, 2019 at 22:04
  • $\begingroup$ @scobaco: Yup, good point, I have edited my answer accordingly. $\endgroup$
    – Sorin Tirc
    Commented Mar 3, 2019 at 22:30
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    $\begingroup$ I do not see how uniform equicontinuity holds given the assumptions. For $g_x(y) = f(x+y) - f(x)$ we have $|g_x(y) - g_x(y_0)| < \epsilon \iff |f(x+y) - f(x+y_0)| < \epsilon|$. For fixed $x$, the mapping $y \mapsto f(x+y)$is uniformly continuous when $y$ is in a compact set, but $x+y$ does not stay in any compact set as $x \to \infty$. $\endgroup$
    – scobaco
    Commented Jan 6, 2020 at 18:00

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