= One-sided approximate identity
{title2=$\int_0^b f(x)k_n(x)\,dx\to f(0)$}
Let nonnegative <integrable functions> $k_n$ on $[0,b]$ satisfy $\int_0^b k_n\to1$ and $\int_\delta^b k_n\to0$ 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 $2\|f\|_\infty\int_\delta^b k_n$. 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 $k_n(x)=2ne^{-n^2x^2}/\sqrt\pi$ 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 $1/2$.
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