 One-sided approximate identity

Let nonnegative integrable functions $k_n$ on $[0,b]$ satisfy ${\int }_0^b{k}_n\to 1$ and ${\int }_{\delta }^b{k}_n\to 0$ for every $0<\delta <b$. If $f$ is a bounded function and continuous at zero, and the products are integrable, then ${\int }_0^{b}f(x)k_n(x)dx\to f(0)$. To prove this, subtract $f(0)\int k_n$ and split the difference integral at $\delta$. The first piece is bounded by ${\sup }_{[0,\delta ]}|f-f(0)|\int k_n$; the second by $2\Vert f{\Vert }_{\infty }{\int }_{\delta }^b{k}_n$. First use continuity to choose $\delta$, then pass to the limit in $n$. The normalization term $f(0)(\int k_n-1)$ also vanishes.

For $b=1$, the Gaussian integral shows that $k_n(x)=2n{e}^{-n^2{x}^2}/\sqrt{\pi }$ has these properties. The factor two compensates for retaining only the nonnegative half of the full normalized Gaussian profile. Replacing it by $n/\sqrt{\pi }$ would instead give limiting mass $1/2$.