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How can I show, just using the mean value inequality that, if $f: U \rightarrow\mathbb{R}^n$ for $U$ open and convex in $\mathbb{R}^m$, with $0 \in U$ and $f(0) = 0$ and $|f'(x)| < |x|$, $\forall x \in U$ (the norm of $f'(x)$ is the usual norm of linear functionals) then $|f(x)| \leq \dfrac{1}{2}|x|^2$, $\forall x \in U$.

Well, since the line $[0,x]$ between 0 and $x$ is such that $|a| \leq |x|$, $\forall a \in [0,x]$ we then have $|f'(a)| \leq |x|$, $\forall a \in [0,x]$. Thus we can apply the mean value inequality between the points $0$ and $x = 0+x$. We therefore have:

$|f(x)| = |f(x) - f(0)| \leq |x|.|x-0| = |x|^2$

I just can't improve my result and find where the term $\dfrac{1}{2}$ shows up.

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This is tricky without integration. The only approach I can think of using only the MVT is to subdivide the interval and apply the MVT separately on each piece. For example if we split $[0,x]$ into the two subintervals $[0,x/2]$ and $[x/2,x]$, then applying the MVT twice we get $|f(x/2)| \le |x/2|^2$ and $f(x) \le f(x/2) + |x||x/2| \le \frac 3 4 |x|^2.$

If we do this with $n$ subdivisions instead, the MVT on the $k^{\rm th}$ interval gives

$$ |f(kx /n)| \le |f((k-1)x/n)| + k|x/n|^2,$$

so we can write

$$ |f(x)| \le \sum_{k=1}^n \frac{k|x|^2}{n^2}.$$

Pulling the $|x|^2/n^2$ out of the sum and recognising the triangular number this becomes

$$|f(x)| \le \frac{n(n+1)|x|^2}{2n^2} = \frac{n+1}{n}\frac 12|x|^2.$$

Since this is true for every $n \in \mathbb N$ and $\frac{n+1}{n} \to 1$, we get the desired bound.

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    Good answer!.... BTW: In the case $m=n=1$ we can show that $f'$ is continuous .(A bit tricky). And therefore $|f(x)|=|f(x)-f(0)|=|\int_0^x f'(y)dy|\leq \int_0^x|f'(y)|dy\leq x^2/2.$ Can this be useful in a method of proof in the general case? – 2017-02-07
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    @user254665: I think you can reduce to the $m=n=1$ case reasonably easily: for fixed $x_0$, restrict the domain of $f$ to the line through $0$ and $x_0$ and then project the image of $f$ on to the line through $0$ and $f(x_0)$. Since this projection is a subisometry the resulting function $[0,|x|] \to \mathbb R$ should also satisfy $|f'(x)|\le |x|.$ If you really can prove that $f \in C^1$ in this simplified case then the integration argument is quite nice - it was certainly my first thought to solve this problem. – 2017-02-07
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    @user254665 Are you claiming that $|f'(x)|\le |x|$ implies $f$ is $C^1?$ – 2017-02-07