Inequality on Right Derivatives of functions $\mathbb{R} \to \mathbb{R}^n$

Question

Let $f: [a,b] \to \Bbb{R}^n$ and $g: [a,b] \to \Bbb{R}$ continuous functions with right-derivative in $(a,b)$. Prove that, if $\left|f'_{+}(x)\right| \leq g'_{+}(x) $, for all $x \in (a,b)$, then $\left|f(b) - f(a)\right| \leq g(b)-g(a) $.

Ok, I have tried a lot, using even parts of the MVT proof theorem, but as $f$ is not differentiable (at least not by hypotessis, if it is and there is some way to prove it, I don't know) at $(a,b)$, I can't proceed to use a lemma. If somebody has some idea in how to proceed (or have some another simple proof) I will be glad! Thanks in advance!

Answer 1

We can reduce the proof to the case $n=1$ as follows: let $\sum |c_i|^{2}=1$ and define $f_0(x)=\sum c_if_i(x)$ where $f=(f_1,f_2,..,f_n)$. Apply the result to the case $n=1$ to the functions $f_0$ and $g$ and take the sup over all $(c_i)$. Note that $|f(b)-f(a)|$ is the supremum of $\sum c_if_i(b)-\sum c_i f_i(a)$ overal $(c_i)$ with $\sum |c_i|^{2}=1$.

As a second step let us reduce the proof to the case when $|f'(x)| <g'(x)$ for all $x$. To do this all you have to do is replace $g$ by $g(x)+\epsilon e^{x}$ to get $f(b)-f(a) \leq g(b)-g(a)+\epsilon (e^{b}-e^{a})$(from the special case) and then let $\epsilon \to 0$.

Suppose $f(b)-f(a) >g(b) -g(a)$. Let $c=\inf \{x: f(x)-f(a) >g(x) -g(a)\}$. There is a sequence $x_n$ decreasing to $c$ such that $f(x_n)-f(a) >g(x_n) -g(a)$ for all $n$. Divide by $x_n-a$ and take the limit to get $f'(x+) \geq g'(x+)$ which is a contradiction.

Finally note that we can replace $f$ by $-f$ to get the inequality $-(f(b)-f(a)) \leq g(b)-g(a)$ so we get $|f(b)-f(a)| \leq g(b)-g(a)$