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Show that there is no distribution $f \in D'(\mathbb{R})$ such that \begin{equation} f(\phi)=\int e^{1/x^2}\phi(x)dx \end{equation} for every $\phi \in C_{0}^{\infty}(\mathbb{R})$ with $supp(\phi) \subset \mathbb{R}-\{0\}$. Thank you.

I actually found a sequence of test functions $\psi_{n}=e^{-1/(1-x)(x-1/n)}(1-x)^n$ such that $f(\psi_n)$ do not tend to $0$, but I don't know how to prove that $\psi_n$ tend to $0$ in $D(\mathbb{R})$, i.e. in the test functions sense (clearly the boundedness of the supports is not the problem, but the uniform convergence of all derivatives is). Clearly, $\psi_n$ tend to $0$ a.e., but this is not enough.

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I think you have to change $u$ by $f$. – Tomás Jan 30 '13 at 10:53
Frank, you function $\psi_n$ seems to be not continuous on $x=1$ and $x=\frac{1}{n}$. – Tomás Jan 30 '13 at 11:37
Thanks, I edited now. – Frank Zermelo Jan 30 '13 at 21:07
On a compact set, a distribution is always of finite order, i.e. you need only the uniform convergence of the derivatives upto the finite order. That's my argument below. – Vobo Jan 30 '13 at 22:44
up vote 1 down vote accepted

First, take a positive test function $\varphi$ that is supported on $(1,2)$ and define for all $j \in \mathbb{N}$, $\varphi_{j}(x)=e^{-j}\varphi(jx)$, which is clearly supported on $(\frac{1}{j},\frac{2}{j})$. It is clear that $\varphi_{j} \in C_{0}^{\infty}(\mathbb{R})$, $supp(\varphi_{j} \subset [0,1]$, for all $j$, $\varphi_{j}(x) \rightarrow 0$ a.e., and also \begin{equation} \frac{d^{i}\varphi_{j}}{dx^{i}}(x)=e^{-j}j^{i}\frac{d^{i}\varphi}{dx^{i}}(jx) \end{equation} therefore \begin{equation} sup_{x \in \mathbb{R}}|\frac{d^{i}\varphi_{j}}{dx^{i}}(x)|\leq e^{-j}j^{i}sup_{x \in \mathbb{R}}|\frac{d^{i}\varphi}{dx^{i}}(x)|\rightarrow 0 \mbox{ a.e. as }j \rightarrow \infty \end{equation} hence $\varphi_{j}$ converge to $0$ in $D(\mathbb{R})$. Now $\int e^{\frac{1}{x^2}} \varphi_{j}(x)dx$ should converge to $0$, but \begin{equation} \int e^{\frac{1}{x^2}} \varphi_{j}(x)dx=\int_{\frac{1}{j}}^{\frac{2}{j}} e^{\frac{1}{x^2}}e^{-j} \varphi(jx)dx=\int_{1}^{2} e^{\frac{j^2}{x^2}}e^{-j}\frac{1}{j} \varphi(x)dx \geq \int_{1}^{2} \varphi(x)dx=||\varphi||_{L^{1}} \end{equation} thus $||\varphi||_{L^{1}}=0$, so $\varphi=0$, contradiction. (above, we used that $j^2/x^2-j-ln(j) \geq j^2/4-j-ln(j)\geq 0$, if $j \geq 6$)

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Assuming a distribution $f\in D'(\mathbb{R}$, for the compact subset $K=[-2,2]$ there is some $C>0$ and some integer $k$ such that $$ |f(\varphi)| \leq C \max_{i\leq k} \max_{x\in K} |D^{(i)}\varphi| $$ for all $\varphi\in C_c^\infty(\mathbb{R})$ with $\operatorname{supp}\varphi\in K$. Choose $\varphi_n\in C_c^\infty(\mathbb{R})$ non-negative with support in $K$, $||\varphi_n||_\infty \leq 1/n^{2k}$ and $\varphi_n(x)=1/n^{2k}$ for $1/n\leq x \leq 2/n$. Then you have $$ f(\varphi_n)=\int_\Omega e^{1/x^2} \varphi_n(x) dx \geq \int_{1/n}^{2/n} e^{1/x^2} \varphi_n(x) dx \geq \frac{1}{n^{2k+1}} e^{n^2/4} \to\infty $$ as $n\to\infty$. You can verify that upto the finite derivation order $k$, the $\varphi_n$ could be choosen bounded, i.e. there is a $B>0$ with $\max_{i\leq k} \max_{x\in K} |D^{(i)}\varphi_n| < B$ for all $n$. This is a contradiction.

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I don't see exactly why we can choose the test functions to satisfy the inequality $max_{i \leq k} max_{x \in K} |D^{(i)}\varphi_{n}| < B$ for all $n$. I didn't find any example for this. However, I've included my proof below, which doesn't use that inequality. – Frank Zermelo Jan 31 '13 at 10:33
You can do it basically like you constructed your function. Nice explicity example you gave. – Vobo Feb 1 '13 at 9:33

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