How to prove that $x \rightarrow e^{1/x}$ is not a restriction of any real distribution to $ \mathbb {R}_+$? This is an excercise 2.2 from Hormander, vol. I:

Does there exist a distribution $u$ on $\mathbb{R}$ with the restriction
$x \rightarrow e^{1/x}$ to $\mathbb{R}_+$?

The answer, provided in the book, is "No". I am trying to "cook up" appropriate test function(s) such that $ \int \phi(x)e^{1/x} \leq C\sum_{\alpha \leq k} \sup\left|\partial^{\alpha}\phi\right|$ for no $k$, and I'm not sure at all what function(s) to take. What is the appropriate function? Is there a general method to come up with just right test functions?
Following Fedja's hint, let $\phi$ be a nonnegative test function supported in $(1,2)$ such that $\phi=1$ on $(5/4,7/4)$. For $b&gt;0$, let $\phi_{b}(x):=\phi(bx)$. Observe that

$$\int_{\mathbb{R}}\phi_{b}(x)e^{1/x}dx=b^{-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx,\qquad\forall b&gt;0$$

Suppose that there is a distribution $u\in\mathcal{D}'(\mathbb{R})$ of order $k$ whose restriction to $\mathbb{R}^{+}$ is $e^{1/x}$:

$$\left|\langle{u,\psi}\rangle\right|\leq C\sum_{\alpha\leq k}\left\|\partial^{\alpha}\psi\right\|_{\infty},\qquad\forall\psi\in C_{c}^{\infty}(\mathbb{R})$$

and in particular,
$$\left|\langle{u,\phi_{b}}\rangle\right|\leq C\sum_{\alpha\leq k}\left\|\partial^{\alpha}\phi_{b}\right\|_{\infty}\leq C\sum_{\alpha\leq k}b^{\alpha}\left\|\partial^{\alpha}\phi\right\|_{\infty}\leq Cb^{k}\sum_{\alpha\leq k}\left\|\partial^{\alpha}\phi\right\|_{\infty},\qquad\forall b\geq 1$$
You can check that there is a constant $C'&gt;0$ such that

$$\dfrac{e^{b/x}}{b^{k+1}}\geq C'\dfrac{b}{x^{k+2}},\qquad\forall x&gt;0$$

Whence,

$$b^{-k}\left|\langle{u,\phi_{b}}\rangle\right|=b^{-k-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx\geq C' b\int_{1}^{2}\phi(x)x^{-k-2}dx$$

Since the RHS tends to $\infty$ as $b\rightarrow\infty$, we obtain a contradiction.
I think the argument is simpler than you expect, unless I make a terrible mistake...

Just pick any test function with the property that $\phi(0) &gt;1$ ($\phi(0)\neq 0$ is actually enough). 

Since $\phi(0)&gt;1$ there exists some $a$ so that $\phi &gt;1$ on $[0,1]$.

Then

$$\int \phi(x) e^{\frac{1}{x}}&gt; \int_0^a e^{\frac{1}{x}} dx = \int_\frac{1}{a}^\infty \frac{e^u}{u^2}du = \infty $$