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Let for $j\in\mathbb N$, $f_j \in \mathcal D'(R^n)$ (distribution) and $f_j$ is locally integrable for all $j$. If $f_j \to f$ in the sense of distribution, is it necessarily true that $f$ is locally integrable?

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2 Answers

up vote 6 down vote accepted

Take $f_k:=k^n\phi(kx)$ where $\phi$ is a nonnegative bump function, whose support is contained in $\left[-1,1\right]^n$, and the integral is $1$. Each $f_k$ is bounded hence locally integrable. We get that $\{f_k\}$ converges to $\delta_0$ in $\mathcal D'(\mathbb R^n)$. Indeed, let $\varphi\in\mathcal D(\mathbb R^n)$. We have \begin{align*} \left|\int_{\mathbb R^n}f_k(x)\varphi(x)dx-\varphi(0)\right|&=\left|\int_{\mathbb R^n}k^n\phi(kx)\varphi(x)dx)-\varphi(0)\right|\\ &=\left|\int_{\left[-\frac 1k,\frac 1k\right]^n}k^n\phi(kx)\varphi(x)dx-\varphi(0)\right|\\ &=\left|\int_{\left[-1,1\right]^n}\phi(y)\varphi\left(\frac yk\right)dy-\varphi(0)\right|\\ &\leq \int_{\left[-1,1\right]^n}\phi(y)\left|\varphi\left(\frac yk\right)- \varphi(0)\right|dy. \end{align*} Since the function $\varphi$ is uniformly continuous on $\left[-1,1\right]^n$, given $\varepsilon>0$ we can find $k_0$ such that for all $k\geq k_0$ and $y\in\left[-1,1\right]^n$ we have $\left|\varphi\left(\frac yk\right)-\varphi(0)\right|\leq \varepsilon$, hence $\left|\int_{\mathbb R^n}f_k(x)\varphi(x)dx-\varphi(0)\right|\leq\varepsilon$. But it's well-known that $\delta_0$ cannot be represented by a locally integrable function (if it was the cases, and $f$ represents $\delta_0$, then take $\phi_k\in\mathcal D(\mathbb R^n)$ such that $\phi_k(x)=1$ if $\lVert x\rVert\leq \frac 1k$, and $0\leq\phi_k\leq 1$. Then $1=\int_{\mathbb R^n}f(x)\phi_k(x)dx$, but thanks to the dominated convergence theorem we can see that the limit of the last integral $k\to\infty$ is $0$).

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+1, nice counterexample. –  Sasha Oct 8 '11 at 12:23
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I have noticed that the heat kernel $H(x,t)$ provides a counterexample too as $t \to 0$, and the proof is similar to the first answer.

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