Continuity of an integral depending on a parameter resembling convolution by mollifiers I am stuck trying to prove/disprove the following:
Let $g(x,y)=\frac{1}{x}\int_{\mathbb{R}}\phi(\frac{y-t}{x})f(t)dt$ where $(x,y)\in (0,\infty)\times\mathbb{R}$.
Here, $\phi\in C_{c}^{1}(\mathbb{R})$ and $f\in C_{c}(\mathbb{R})$.
The problem asks if $\lim_{(x,y)\rightarrow(0,y_{0})}g(x,y)$ exist with some fixed $y_{0}\in\mathbb{R}$.
I am aware of the standard application of the Dominated Convergence Theorem which would give the continuity of such integrals in $(0,\infty)\times\mathbb{R}$. However, I think the issue at $x=0$ means that that wouldn't work.
If $\phi$ were a standard mollifier, I think $\frac{1}{x}\phi(\frac{y-t}{x})$ approaches $\delta(y-t)$ hence the limit of $g$ would be $f(y_{0})$. So I'm looking for counterexamples with less regular $\phi$, but haven't found any yet. I would appreciate any help. Many thanks in advance!

If $f$ is continuous and bounded and $\phi $ is $L^1$ and $C = \int_{-\infty}^\infty \phi(u)du$ and $a\in [-A,A]$, as $n\to\infty$
$$|f(a)-\int_{-\infty}^\infty n \phi(n(t-a))f(t)dt| =|\int_{-\infty}^\infty  \phi(u)(f(\frac{u}{n}+a)-f(a))du|$$
$$\le \int_{|u|\ge n^{1/2}} | \phi(u)(f(\frac{u}{n}+a)-f(a))du|+\int_{|u|\le n^{1/2}} | \phi(u)(f(\frac{u}{n}+a)-f(a))du|$$
$$\le \|f\|_\infty \|\phi 1_{|u|>n^{1/2}}\|_{L^1}+\|\phi\|_{L^1} \sup_{|u-v|\le n^{-1/2}, |u|\le 2A} |f(u)-f(v)| $$
