I am learning the distribution theory currently. My professor in class said that,

if we defined $J_1(\varphi):=\sum_{n=0}^\infty \varphi^{(n)}(0)$ and $J_2(\varphi):=\sum_{n=0}^\infty \varphi^{(n)}(n)$, where $\varphi\in C^\infty_0(\mathbb R)$, then both $J_1$ and $J_2$ were not distributions(i.e. $J_1$ and $J_2$ were not continuous linear functional on $C^\infty_0(\mathbb R)$).

Clearly $J_1$ and $J_2$ are linear. Suppose $\varphi_k\to0$ in $C^\infty_0(\mathbb R).$ Then given any $\epsilon \gt0,$ for every positive integer $i$, we may find an $N_i\in\mathbb N$ such that $sup_{x\in\mathbb R} |\varphi_k^{(i)}(x)| \lt {\epsilon\over{2^i}}$ when $k\ge N_i$. Hence when $k\to \infty$, lim$_{k\to\infty}J_1(\varphi_k)=\text{lim}_{k\to\infty}\sum_{n=0}^\infty \varphi^{(n)}_k(0)=\text{lim}_{k\to\infty}\text{lim}_{m\to\infty}\sum_{n=0}^m \varphi^{(n)}_k(0)=$

$\text{lim}_{m\to\infty}\text{lim}_{k\to\infty}\sum_{n=0}^m \varphi^{(n)}_k(0)=0,$ where we can interchange the limits because the series is absolutely convergent. So $J_1$ is continuous. Similarly we can show that $J_2$ is continuous.

So the only possibility that $J_1$ and $J_2$ are not distributions is that $J_1(\varphi_1)=\infty$ and $J_2(\varphi_2)=\infty$ for some $\varphi_1,\varphi_2\in C^\infty_0(\mathbb R).$

My question is, can someone give me examples of $\varphi_1,\varphi_2\in C^\infty_0(\mathbb R)$, such that $$J_1(\varphi_1)=\sum_{n=0}^\infty \varphi_1^{(n)}(0)=\infty$$ and $$J_2(\varphi_2)=\sum_{n=0}^\infty \varphi_2^{(n)}(n)=\infty.$$

Thanks in advance.

  • $\begingroup$ $J_1$ is not defined. This is because 1) there exists divergent series and 2) any serie can be realized using the derivatives of some smooth functions at a point. the latter is the Borel-Ritt Theorem encyclopediaofmath.org/index.php/Borel_theorem $\endgroup$ – Abdelmalek Abdesselam Feb 4 '20 at 15:08
  • $\begingroup$ Yes, I realized it is not defined after reading @reuns 's post. The theorem you mentioned is so powerful and amazing, and I never know such theorem before. Thanks for the info. $\endgroup$ – Sam Wong Feb 5 '20 at 15:12

$J_2$ is of course a distribution.

For $J_1$, let $\psi \in C^\infty_c, \psi=1$ on $[-1,1]$ then look at $$\lim_{N\to \infty} J_1(\psi \sum_{n=0}^N \frac{x^n}{n!})$$

  • $\begingroup$ Oh yea, $J_2$ must be a distribution. And your example for $J_1$ is nice. But can you tell me what the philosophy behind this construction is? i.e. How can we think of this construction. Thanks! $\endgroup$ – Sam Wong Feb 2 '20 at 12:39
  • 1
    $\begingroup$ The philosophy is that your functional has infinite order : knowing that $\sum_{k\le n} \|\phi^{(k)}\|_\infty$ is small doesn't imply that $J_1(\phi)$ is small, thus we can construct a sequence such that $\phi_n\in C^\infty_c(-r,r)$, $J_1(\phi_n)=1$ and $\sum_{k\le n} \|\phi_n^{(k)}\|_\infty\le 2^{-n}$ which implies that $\sum_n \phi_n$ converges in $C^\infty_c$ while $J_1(\sum_n\phi_n)=\infty$ $\endgroup$ – reuns Feb 4 '20 at 4:58

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