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Consider the Dirac delta distribution $\delta$ in $\mathbf{R}^d$. It is quite standard to approximate it by functions $g_n$ with $\|g_n\|_{L^1} = 1$.

Is it possible to choose a sequence of test functions $\{g_n\}$ converging to $\delta$ as a distribution such that their derivatives are $L^1$-bounded? I mean, such that for a given a natural number $k$ there is a constant $M>0$ so that

$\|\partial^\mu g_n \|_{L^1} \le M$

for all $n$'s and all multi-indices $|\mu| \le k$?

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up vote 2 down vote accepted

As paul garrett commented, we can employ integration by parts. Let $g_n$ be a sequence of functions tending, as distributions, to the dirac $\delta$. Integration by parts yields $$ \lim_{n\to\infty}\int g_n^\prime(x)\,f(x/\epsilon)\,\mathrm{d}x=-f'(0)/\epsilon\tag{1} $$ Note that $\|f(x/\epsilon)\|_{L^\infty}=\|f(x)\|_{L^\infty}$. Therefore, Hölder's Inequality says $$ \lim_{n\to\infty}\|g_n'\|_{L^1}\|f\|_{L^\infty}\ge\left|\,\lim_{n\to\infty}\int g_n^\prime(x)\,f(x/\epsilon)\,\mathrm{d}x\,\right|=|f'(0)/\epsilon|\tag{2} $$ Since $\epsilon$ is arbitrary, we get that $$ \lim_{n\to\infty}\|g_n'\|_{L^1}=\infty\tag{3} $$

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You might want to quote @paulgarrett's comment to this effect. – Did Jan 20 '13 at 10:55
I see that. Thanks. – robjohn Jan 20 '13 at 11:03
thanks for the very well written answer! – Yul Otani Jan 21 '13 at 13:36

No, the whole point of the delta function is that it is very tall and very narrow, so the derivative is tall/narrow which blows up.

More explicitly, take $\mathbb R^1$ and the triangle approximation. The base is $(\frac{-1}n,\frac 1n)$ and the peak is $(0,n)$ so the slope is $n^2$. Other versions will be similar.

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And the inevitability of the similarity of any example to the explicit (successful) ones is by integration by parts. – paul garrett Jan 20 '13 at 1:21
thank you! yeah you're right... i kinda needed that, though :( – Yul Otani Jan 20 '13 at 2:10

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