# Uniform convergence of $\sqrt{x+y} - \sqrt{x}$ on compact sets but not on $\mathbb{R}$

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?

• 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. Commented Mar 1, 2019 at 22:32
• @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. Commented Mar 3, 2019 at 21:53
• Oh, I see. I got confused with his wording. Commented Mar 3, 2019 at 23:23

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 $$0, by the triangle inequality we get $$\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 $$A_n(\varepsilon)=\emptyset$$.

• 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? Commented Jan 8, 2020 at 17:40

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

• Look, maybe I am confused, maybe you are. You were asking something about compact sets, have you changed your goal now? Commented Mar 3, 2019 at 22:04
• @scobaco: Yup, good point, I have edited my answer accordingly. Commented Mar 3, 2019 at 22:30
• 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 $x \to \infty$. Commented Jan 6, 2020 at 18:00