In class we are following the book by Hormander called Linear Partial Differential Operators.
It is stated that: Let $u$ be an integrable function and vanish outside a compact subset $K$ of $\Omega$ ( $\Omega$ is an open set in $R^n$ dimensional space), then $u_\epsilon \in C_0^{\infty}(\Omega)$ if $\epsilon$ is smaller than the distance $\delta$ from $K$ to $\Omega^c$.
Where $u_\epsilon := \int_{R^n} u(x-\epsilon y) \phi(y) \ dy$ and $\phi$ is a test function.
To prove the differentiability we wish to bring the limit as $h \rightarrow 0$ inside the integral, I wrote in my notes that the bound (In the case $n = 1$) would be
$$| u(y)\frac{\phi_\epsilon(x+h) - \phi_\epsilon(x)}{h} | \le 2 |u(y)| $$ where $\phi_\epsilon := \epsilon^{-n} \phi(x/ \epsilon) $ allows us to apply the lebesgue dominated convergence theorem to take the limit inside the integral, how was this bound found? A fellow student of mine suggested to me that $\phi_\epsilon \le 1$ but here we have an $h$ in the denominator.
EDIT: The bound that seems more natural to me is $$| u(y)\frac{\phi_\epsilon(x+h) - \phi_\epsilon(x)}{h} | \le |u(y)| \sup_h \{ \frac{\phi_\epsilon(x+h) - \phi_\epsilon(x)}{h}\} $$
and the supremum would exist and be finite since the test functions are continuous and compactly supported (and compositions of continuous functions are continuous).