limit and integral in the Lp space with Schwartz class.

Question

Can we actually interchange limit and integral for $f \in S(\mathbb{R})$. Or it's just the limit and integral happen both equal to 0?

An exercise: Show that we can interchange the limit and integral: $\lim_{y\rightarrow0} \int_{\mathbb{R}} |f(x-y)-f(x)|^p dx = \int_{\mathbb{R}} \lim_{y\rightarrow0} |f(x-y)-f(x)|^p dx = 0,$ for $f \in S(\mathbb{R})$.

This an exercise from a book, I showed they separately equal to 0 and therefore forms the equal sign

Answer 1

For $|y|\le 1$ and $p > 1/(2n)$

$$\sup_x\ (1+x^{2n}) |f(x)-f(x-y)|\le \sup_x \int_{-y}^0 (1+x^{2n})|f'(x+t)|dt $$ $$\le \sup_x \int_{-y}^0 (1+(x+t+1)^{2n}) |f'(x+t)|dt$$ $$ \le \int_{-y}^0 \sup_x (1+(x+t+1)^{2n}) |f'(x+t)|dt = |y| A$$

where $A = \sup_x (1+(x+1)^{2n}) |f'(x)|$ is finite because $f'$ is Schwartz.

Thus

$$\int_{\mathbb{R}} |f(x-y)-f(x)|^p dx\le \int_{\mathbb{R}} \frac{A|y|}{(1+x^{2n})^p} dx$$