Define $\phi_t(x)=\frac{1}{t\sqrt\pi}e^{-(\frac{x}{t})^2}$ and let $f$ be a bounded and continuous function. It has been shown that $\phi_t(x)$ has properties of a good kernel. I was thinking of showing that the difference $\int_{-\infty}^{\infty} f(x)\phi_t(x)dx - f(0)$ will tend to zero. We can rewrite this using the fact that $\int_{-\infty}^{\infty} \phi_t(x)dx=1$ as $$\int_{-\infty}^{\infty} [f(x)-f(0)]\phi_t(x)dx$$
We know that $\phi_t(x)$ will get large near the origin as $t \rightarrow 0$ and will disappear everywhere else. So then we can break up the limits of integration by taking a tiny $\delta>0$ $$\int_{-\infty}^{\infty} [f(x)-f(0)]\phi_t(x)dx=\int_{-\infty}^{-\delta} [f(x)-f(0)]\phi_t(x)dx+\int_{-\delta}^{\delta} [f(x)-f(0)]\phi_t(x)dx+\int_{\delta}^{\infty} [f(x)-f(0)]\phi_t(x)dx$$
It is easy to see that everything outside of $(-\delta,\delta)$ goes to zero since $f$ is bounded and continuous, so it will not blow up and $\phi_t(x)$ will kill it off. I am struggling showing the little part of the integral where $\phi_t(x)$ blows up will go to zero. It makes sense intuitively since $f(x)-f(0)$ is very small inside our $\delta$ interval, and $\phi_t(x)$ gets highly concentrated at zero, but I am not sure how to rigorously explain this.