Consider a function $g(x,y)$, $x\in X$ and $y\geq 0$ where $X$ is a compact subset of $\mathbb{R}$. Assume that $g(x,y)$ converges pointwise to zero as $y\downarrow 0$, for all $x\in X$.
Is the convergence also uniform (or, under what conditions on $g$ is the convergence uniform)?
I think the answer is "yes", but I have some difficulties in a step of the proof. Here is my approach.
By assumption of pointwise convergence, for all $\varepsilon >0$ and for all $x\in X$ there exists a $\delta (x,\varepsilon)$ such that $|g(x,y)|<\varepsilon$ for all $y<\delta(x,\varepsilon)$. I can choose $\delta(x,\varepsilon)$ such that $|g(x,\delta (x,\varepsilon )|=\varepsilon$. If I can prove that $\delta(x,\varepsilon)$ is continuous in $x$, then there exists a $\delta^* (\varepsilon) = \min_{x\in X} \delta(x,\varepsilon)$ (because I am taking the min of a continuous function on a compact space) such that $|g(x,y)|<\varepsilon$ for all $y<\delta^* (\varepsilon)$ and for all $x\in X$, which would complete the proof.
However, I am not sure how to formally show that $\delta(x,\varepsilon)$ is continuous. I can add continuity assumptions on $g$, if necessary, but I cannot assume $g$ is monotone (that is, I cannot invoke Dini's Theorem).
You need to add some assumptions on $g$; otherwise a counterexample would be $$ X=[0,1]\qquad g(x,y)=\begin{cases}0& x=0 \\ y/x & x\ne 0 \end{cases}$$
However, if $g$ is continuous, then for any $\varepsilon>0$, the set $g^{-1}((-\varepsilon,\varepsilon))$ is open as a subset of $X\times \mathbb R$ and contains the $x$-axis. At each point on the $x$-axis we can choose a nontrivial ball that is a subset of this preimage, and the radii can be chosen to vary continuously. Since $X$ is compact, there is a minimum radius, which can be used as a uniform $\delta$.
(Also, if $g$ is continuous, the pointwise limit doesn't need to be specified explicitly, just that $g(x,0)=0$ for all $x$).