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For integrable functions defined on the real line, the normalized gaussian function approximates the convolution identity, Dirac Delta, in the sense that if $$g(t):=N_0e^{-x²}$$ (denoting the normalizing constant by $N_0$) and $$g_{a0}(x):=aN_0e^{-(ax)²}$$ then, for all $f$ integrable, $$\lim_{a\to\infty} f*g_{a0}=f$$ almost everywhere.

My first question is: Is this true?

My second question is: does the $n$-th derivative of the Gaussian function approximates the derivade operator by convolution? I mean writing $$g_{an}:=\frac{d^n}{dx^n} g_{a0}(x)$$ then $$\lim_{a\to\infty} f*g_{an}=f^{(n)}$$ and (third question) whether the $n$-th derivative operator includes or not the normalization constant $N_n$ for $g_{an}$.

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