Proving $e^{1/x}$ is not a distribution. Let $\mathcal{S}(\mathbb{R})$ denote the space of Schwartz functions and $\mathcal{S}'(\mathbb{R})$ its dual, the space of tempered distributions. By definition, any $u\in\mathcal{S}'$ satisfies that $|u(\phi)| \le C\sup_{x, a\le N, b\le M} |x^a\phi^{(b)}(x)|$ for some constant $C>0$ and some integers $N,M\ge 0$. How could we prove that $e^{1/x}$ is not the restriction to $(0,\infty)$ of a tempered distribution, i.e. that there exists no $u\in\mathcal{S}'$ such that $$ u(\phi) = \int_{\mathbb{R}} e^{1/x}\phi(x) dx \hspace{1cm}\forall \phi\in C^\infty_c((0,\infty))?$$
My idea is to focus only on $\phi\in C^\infty((0,1))$ real-valued and non-negative, so that $u(\phi) \le C\sup_{x\in [0,1],b\le M} |\phi^{(b)}(x)|$. Given any $\epsilon>0$, I think we can find a bump function $\phi$ with support in $(\epsilon,1-\epsilon)$ such that it is identically $1$ on $(2\epsilon, 1-2\epsilon)$. Moreover, we can ask for $|\phi'| \le 2/\epsilon$, since the bump function has to go from $0$ to $1$ on $[\epsilon,2\epsilon]$. Similarly, we should be able to ask for $|\phi^{(b)}(x)| \le (4/\epsilon)^b$ for all $b=0,1,\ldots,M$, but I'm having trouble rigorously proving such a bump function exists. Assuming it does, however, we get
$$ \int_{2\epsilon}^{1-2\epsilon} e^{1/x} dx \le C\left(\frac{4}{\epsilon}\right)^M, $$
or equivalently $\int_{\delta}^{1-\delta}e^{1/x}dx\le C(8/\delta)^M$ for all $\delta>0$. Then we can easily finish by noticing $e^{1/x}\ge \frac{(1/x)^k}{k!}$ for all $k$ and using $k=M+2$, and letting $\delta\to 0$. My question is: is my procedure to find the bump function valid?
 A: Something like that should work. What occurred to me was choosing $\phi\in C^\infty_c$ with support $[1/2,1]$ such that $\phi>0$ in $(1/2,1).$ Then for $0<\epsilon<1,$ define $\phi_\epsilon(x)=\phi(x/\epsilon).$ We then have $\phi_\epsilon$ supported in $[\epsilon/2,\epsilon].$ Note that $D^m\phi_\epsilon (x) = \epsilon^{-m}D^m\phi(x/\epsilon).$
Suppose there exists $u\in S'$ such that
$$\tag 1 u(\phi_\epsilon)=\int_{\epsilon/2}^\epsilon e^{1/x}\phi_\epsilon(x)\,dx$$
for all $\epsilon.$ Then there are constants $C,M$ such that
$$\tag 2 |u(\phi_\epsilon)|\le C\sup_{x\in[0,1],m\in \{0,1,\dots,M\}}|D^m\phi_\epsilon(x)|$$
for all $\epsilon \in (0,1).$ The right side of $(2)$ is no more than $C_1\epsilon^{-M}.$ But the right side of $(1)$ equals
$$\epsilon\int_{1/2}^1 e^{1/(\epsilon y)}\phi(y)\,dy \ge \epsilon\cdot e^{1/\epsilon }\int_{1/2}^1 \phi(x)\,dx.$$
Since $\epsilon^{M+1} e^{1/\epsilon }\to \infty$ as $\epsilon\to 0^+,$ we have a contradiction.
A: Assuming $supp f\subset (0,1)$ we get $|u(f)|\le C\sup_{b,x}|f^{(b)}(x)|$. Assume we can construct a smooth $f:\mathbb{R}\to [0,1]$ such that $supp f\subset [\epsilon, 1-\epsilon]$ and $f=1$ on $[2\epsilon,1-2\epsilon]$. Then we get 
$$ \int_{2\epsilon}^{1-2\epsilon} e^{1/x} dx \le \int e^{1/x}f(x) dx \le C\sup_{b,x}|f^{(b)}(x)|.$$
We construct $f$ symmetrically about $1/2$, so we need only worry about defining it on $[\epsilon,2\epsilon]$, and moreover the derivatives of $f$ are 0 away from $[\epsilon,2\epsilon]\cup [1-2\epsilon,2\epsilon]$, so by symmetry we need only take the above $\sup$ on $x\in [\epsilon,2\epsilon]$. Pick any smooth $g: \mathbb{R}\to [0,1]$ such that $g=0$ on $(-\infty,0]$ and $g=1$ on $[1,\infty)$. Then define $f(x) = g\left(\frac{x}{\epsilon}-1\right)$ for $x\in [\epsilon,2\epsilon]$. Then $f^{(b)}(x) = \frac{1}{\epsilon^b}g^{(b)}\left(\frac{x}{\epsilon}-1\right)$. Let $D:=\sup_{x\in[0,1],b} |g^{(b)}|<\infty$ since $0\le b\le M$ has finitely many choices and $[0,1]$ is compact and all the derivatives of $g$ are continuous on $[0,1]$. Then $\sup_{b,x}|f^{(b)}(x)| \le \frac{D}{\epsilon^M}$. Finally, recall $e^{1/x}=(1/x) + (1/x)^2/2!+\cdots$, so $e^{1/x}\ge \frac{1}{k!x^k}$ for any $k\ge 0$ integer; choose $k=M+2$. Then the above inequality becomes:
$$\frac{CD}{\epsilon^M} \ge \frac{1}{k!}\int_{2\epsilon}^{1-2\epsilon} x^{-k} dx = \frac{1}{k!}\left(\frac{1}{1-k}x^{1-k}\right)^{x=1-2\epsilon}_{x=2\epsilon} = \frac{\frac{1}{(2\epsilon)^{k-1}}-\frac{1}{(1-2\epsilon)^{k-1}}}{k!(k-1)}. $$
Letting $\delta=2\epsilon$ and incorporating all the constant into one $C'$, we get
$$ C' \ge (\delta^M)\left(\frac{1}{\delta^{k-1}}-\frac{1}{(1-\delta)^{k-1}}\right)\hspace{.8cm}\forall 0<\delta<0.1.$$
Notice $1/\delta^{k-1} \ge 2/(1-\delta)^{k-1}$ for small enough $\delta$, since $\left(\frac{1}{\delta}-1\right)^{k-1}\ge 2$ for small enough $\delta$, since $k-1\ge 1$. Thus the above inequality implies $2C' \ge \frac{\delta^M}{\delta^{k-1}}$ for small enough $\delta>0$, i.e. $2C'\ge \delta^{M+1-k}=\frac{1}{\delta}$, our desired contradiction. 
