# $f \in \mathcal S, f(0)=1$ then $\lim_{\epsilon \to 0} f(\epsilon x) = 1$

Let $f \in \mathcal S(\mathbb R^n)$ with $f(0) = 1$. Here $\mathcal S$ means the Schwartz class. Then how can I prove that $$\lim_{\epsilon \downarrow 0} f(\epsilon x) = 1 \; \text{(compact convergence)} \;?$$

-
As $\sup_{x\in\Bbb R}|f(x\varepsilon)-1|=\sup_{t\in\Bbb R}|f(t)-1|$, I don't how there could be uniform convergence on the real line. But the is convergence on all the compact subsets. – Davide Giraudo Jan 31 '13 at 12:47
@DavideGiraudo Thank you for the comment. I edited my question. :) – Konsta Jan 31 '13 at 13:07
$|f(\epsilon x)-f(0)| \leq \sup |f'| \cdot \epsilon |x|$, and if $x$ tanges over a bounded subset, then... – Siminore Jan 31 '13 at 13:19

Let $[-R,R]^d$ a compact subset of $\Bbb R^d$. Let $g(x,t):=f(tx)$. $$f(\varepsilon x)-1=g(x,\varepsilon)-g(0,\varepsilon)=\int_0^{\varepsilon}\partial_tg(x,t)dt=\sum_{j=1}^d\int_0^\varepsilon\partial_jf(x,t)x_j dt,$$ which gives for $\varepsilon<1$ that $$|f(\varepsilon x)-1|\leqslant R\cdot d\cdot\varepsilon\max_{1\leqslant j\leqslant d}\sup_{\substack{|x|\leqslant R\\ |t|\leqslant 1}}|\partial_jf(x,t)|.$$ The latest supremum is finite as $f$ is $C^1$ (so we don't need $f$ to be in the Schwarz space).