I need to show that this inequality holds:
$| \int_{0}^{1} (h \nabla f(x+sh-y) -h \nabla f(x-y)) ds | \leq |h| \varepsilon(|h|)$
For a function $\varepsilon$ which verifies $\varepsilon (|h|) \to 0 $ as $|h| \to 0$, where $f \in C_{c}^{1}(\mathbb{R}^n)$, and $h \in \mathbb{R}^n$ with $|h|<1$.
This is supposed to be obtained from the fact that $\nabla f$ is uniformly continuous on $ \mathbb{R}^n$.
I don't know how to proceed, I know that if a continuous function f is compactly supported, then it is uniformly continuous. As $f \in C_{c}^{1}(\mathbb{R}^n)$, $\nabla f$ is also compactly suported and continuous, so it is also uniformly continuous.
But I don't know how to pass it to the given integral.
Edit:
Following Andrew D. Hwang's clue, we have:
$| \int_{0}^{1} (h \nabla f(x+sh-y) -h \nabla f(x-y)) ds | \leq \int_{0}^{1} |(h \nabla f(x+sh-y) -h \nabla f(x-y))| ds \leq \int_{0}^{1} |(h \nabla f(x+sh-y)| +|h \nabla f(x-y))| ds$.
Here, as the first part of the integral is equal to $ f(x+h-y)-f(x-y)$, and f is uniformly continuous, we have that for any $x,y \in \mathbb{R}^n$ there exists $\varepsilon_1$ such that $\varepsilon_1(|h|) \to 0$ as $|h| \to 0$ so that this inequality holds:
$$\int_{0}^{1} |(h \nabla f(x+sh-y)| \leq \varepsilon_1(|h|) $$
And argue the same way for the second part of the integral, which verifies $\int_{0}^{1}|-h \nabla f(x-y)|ds=|-h \nabla f(x-y)|$, obtaining $\varepsilon_2$.
Then $\varepsilon= \varepsilon_1+\varepsilon_2$. Is this correct?
Note: This inequality is needed in the book Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis, in the proof of the proposition 4.20.
Let $\epsilon(h)= \sup_{|z-w|\le |h|}|\nabla f(z) - \nabla f(w)|.$ By the uniform continuity of $\nabla f,$ $\lim_{h\to 0} \epsilon(h) = 0.$ We then have
$$| \int_{0}^{1} (h \nabla f(x+sh-y) -h \nabla f(x-y))\, ds | \le \int_{0}^{1} |h |\,|\nabla f(x+sh-y) -\nabla f(x-y)| \,ds$$ $$ \le \int_{0}^{1} |h |\,\epsilon(h) \,ds = |h |\,\epsilon(h)$$
as desired.