$e^x $ is not a tempered distribution

[Camillo]

Prove that $e^x$ is not a tempered distribution .

Hint

The definition of tempered distribution is :
A distribution $u in ccD'(RR) $ is named tempered when it verifies the condition: $lim_(k rarr oo) int_(RR) uv_k =0 $
for each sequence $(v_k) $ of elements $ in ccD(RR)$ which converges to $0 $ in the meaning of $ccS $(Schwartz space i.e. space of regular functions rapidly decreasing).
The space of tempered distributions on $ RR $ is called $ccS' (RR)$.

[Camillo]

"Kroldar":I think that a different way to prove that $e^x$ is not a tempered distribution is by considering that $e^(-|x|) in ccS$ and $e^x$ is not defined on $e^(-|x|)$.

You mean ...and $ e^x $ is not defined on $ccS$ ?
Pls explain what you mean

[Kroldar]

I think that a different way to prove that $e^x$ is not a tempered distribution is by considering that $e^(-|x|) in ccS$ and $e^x$ is not defined on $e^(-|x|)$.

[Camillo]

Here is "my" solution.

Let's fix a function $v in ccD(RR) $ non negative and such that $v =1$ in $[0,1] $ ; let's consider the sequence of functions $v_k(x)=(v(x/k))/2^k $.

It is easy to prove that $v_k rarr 0 $ in $ccS $.
We have consequently:
$int_(-00)^(+oo) e^xv_k(x) dx=k/2^k int_(-oo)^(+oo)e^(ky)v(y)dy >=k/2^k int_0^1e^(ky) dy=(e^k-1)/2^k $.

The sequence of integrals is not infinitesimal and therefore $e^x $ does not belong to $ccS'$.