Analysis problem: Show that$ f(x)$ is less or equal than$ g(x)$

Question

Analysis problem:

Let $f$ and $g$ be differentiable on $ \mathbb R$. Suppose that $f(0)=g(0)$ and that $f' (x)$ is less or equal than $g' (x)$ for all $x$ greater or equal than $0$ Show that $f(x)$ is less or equal than$g(x)$ for all $x$ greater or equal than $0$.

Is my proof correct?

I am trying to use the Generalized Mean Value Theorem:

As $f$and g are differentiable on$ \mathbb R$, $f$ and $g$ are continuous on $ \mathbb R$ and we can use the Generalized Value Theorem. Using the starting condition $f(0)=g(0)$, we have that for any b that is greater than$ 0$, exist a $c$ element of $(0,b)$ such that

$f' (c) g(b) = g' (c) f(b)$

By the starting conditions,

$f' (x) $is less or equal than$g' (x)$ for all $x$ greater or equal than $0$

Therefore, $f(b)$ is less or equal than $g(b)$ for any b element of $(0, b)$

As $b$ is any number bigger than$ 0$

$f(x)$ is less or equal than $g(x)$ for any $x$ greater or bigger than$ 0$. Q.E.D.

Answer 1

Let $h(x)=f(x)-g(x);x\in [0,\infty)$

$h^{'}(x)=f^{'}(x)-g^{'}(x)\le 0\implies h$ is decreasing on $[0,\infty)\implies h(x)\le h(0)\forall x\in [0,\infty)\implies f(x)\le g(x)$

Answer 2

I don't really see how you conclude from $ f'(c)g(b)=g'(c)f(b) $ and $f'(c)\leq g'(c)$ that $f(b)\leq g(b)$. For instance, $0<2$ and $0\times (-1) = 2\times 0$, but you cannot conclude that $0\leq-1$.

As mentioned by dxiv, a proof can be achieved by using the Mean Value Theorem, applied to the function $g-f$: given any $x\geq0$, there exists $c$ with $0\leq c\leq x$ and $$g(x)-f(x)=g(x)-f(x)-(g(0)-f(0))=(g'(c)-f'(c))\,(x-0)\geq0.$$

It is also worth mentioning that the "direct way": $$ f(x)=\int_0^xf'(t)\,dt\leq\int_0^xg'(t)\,dt=g(x) $$ doesn't work in general, because it is not true in general that $f(x)=\int_0^xf'(t)\,dt$; this requires $f$ to be absolutely continuous (see also here and here).