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?
 A: 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 
$A_n(\varepsilon)=\emptyset$.
A: 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
