How to prove that $\frac d{dx}e^x=e^x$ using it's infinite series? I got $\sum^\infty_{n=0}\frac{x^{n-1}}{(n-1)!}$ so far but I am having trouble saying that it is equal to the series expansion of $e$
Well just so I am understanding this correctly. There is a function $f \in S$ which is multiplied with some polynomial in the form $\frac{1}{x_1^2x_2^2...}$. Now my first question is that what do they mean "outside of a compact set, f is being dominated by..."
@masfenix They mean that there is a compact set $K\subseteq \Bbb R^n$ such that over $\Bbb R^n\setminus K$ your function is dominated by $$\frac{1}{x_1^{2p}\cdots x_n^{2p}}$$ meaning it is less than that. Being positive, it is integrable by comparison.
@PedroTamaroff thank you. okay so let me recap. We want to show that every Schwartz function (a function that decreases rapidly when multiplied by inverse power). Now, consider some function $f \in S$ where S is the Schwartz space. Now clearly, inside some compact set, f is continuous so that argument is easy. But outside the compact support K (ie, consider some ball with radius R), we have that $f$ times $x_1^{2p}...x_n^{2p} < 1$ is so its bounded (correct?)
I admit I'm being stupid, but I'm switching to spherical coordinates. With trig stuff in the denominator, is it obvious there's no problem? @Kevin... You missed an OUT.
@PedroTamaroff I wrote a recap of everything which I think is correct, but I am a bit confused on what the interpretation of the integral you wrote down is
@masfenix The interpretation is that all $f \in S$ decrease so fast at infinity that the area underneath them is bounded, even though the domain is unbounded
and we proved that by finding a function that also has bounded area from a large cube out to infinity and then argued that outside that cube all $f \in S$ are less than the function we found
ahh I see. Also last question and I think I have it then. When you wrote down $x_1^{2p}...x_n^{2p} \cdot f(x) < 1 $ what does that mean? is that what @KevinDriscoll just wrote?
