Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

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 $j \geq 6$)

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.