# Need an example of $C^\infty_0$ functions with some conditions

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 $$J_2(\varphi_2)=\sum_{n=0}^\infty \varphi_2^{(n)}(n)=\infty.$$

• $J_1$ 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 – Abdelmalek Abdesselam Feb 4 '20 at 15:08
$$J_2$$ is of course a distribution.
For $$J_1$$, let $$\psi \in C^\infty_c, \psi=1$$ on $$[-1,1]$$ then look at $$\lim_{N\to \infty} J_1(\psi \sum_{n=0}^N \frac{x^n}{n!})$$
• 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! – Sam Wong Feb 2 '20 at 12:39
• 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 $J_1(\sum_n\phi_n)=\infty$ – reuns Feb 4 '20 at 4:58