Let $f:U\rightarrow \mathbb{R}^{n}$ of class $C^{1}$ in the open and convex set $U\subset R^{m}$, with $0\in U$ and $f(0)=0$. If $|f'(x)|\leq |x|$ for all $x\in U $, then $|f(x)|\leq\frac{1}{2}|x|^{2}$ for all $x\in U$.
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I tried to use the fact that if $U$ is a convex set and $0\in U$, then for all $x\in U$, $\left[ 0,x \right]\subseteq U$, and $\left[ 0,x \right]$ is a compact set in $U$. Therefore, $f$ is uniformly differentiable since $f$ is $C^{1}$. I do not know if I can choose a suitable $\epsilon$ to be able to limit the value of f in x such that the desired inequality is given – Felipe Torres Oct 19 '17 at 16:42
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I think you can prove that $|f(x)| \leq |x|^2 / 2$. – Rigel Oct 19 '17 at 16:58
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yes, I didn't write the square of the norm of x, thanks for the clarification. – Felipe Torres Oct 19 '17 at 17:13
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Let $x \in U$ and $g(t) = f(tx)$. Then $$f(x) = f(x)-f(0) = g(1)-g(0) = \int_0^1 g^{\prime}(t)\, dt = \int_0^1 x\cdot f^{\prime}(tx)\, dt \le \int_0^1 t|x|^2\, dt = |x|^2/2.$$ I am quite sure that's what you really want.
Petr Naryshkin
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Thanks, I miss writing the square of $|x|$ in the last inequality. You are very kind, it helps me a lot. – Felipe Torres Oct 19 '17 at 17:16