inequality of two decreasing functions

Question

Let $f(x)$ and $g(x)$ be continuous functions for all $x \in \Re$. I want to show that $\forall x$ $$g(x) \leq f(x).$$ Here's what I know about $f$ and $g$:,$$f(x_0)=g(x_0)$$ and for $x>x_0$, $$g^{\prime}(x) \leq f^{\prime}(x).$$ Also, for $x<x_0$, $$f^{\prime}(x) \leq g^{\prime}(x).$$ I'm not sure if this is even possible but please let me know. Thank you!

Answer 1

First off, just knowing that $f$ and $g$ are continuous isn't enough for us to be able to talk about $f'$ and $g'$ at all. For that, we need to assume that $f$ and $g$ are differentiable on $\Bbb R.$

I will assume a bit further still, and suppose that $f$ and $g$ are continuously differentiable--that is, that $f'$ and $g'$ are continuous functions on $\Bbb R.$ Consequently, we have for all $x\in\Bbb R$ that $$f(x)=f(x_0)+\int_{x_0}^xf'(t)dt$$ and $$g(x)=g(x_0)+\int_{x_0}^xg'(t)dt.$$

Consequently, we have for all $x\in\Bbb R$ that $$f(x)-g(x)=f(x_0)-g(x_0)+\int_{x_0}^xf'(t)dt-\int_{x_0}^xg'(t)dt.$$ By the first known condition, we have that $$f(x)-g(x)=\int_{x_0}^x\bigl(f'(t)-g'(t)\bigr)dt$$ for all $x\in\Bbb R.$ By the second known condition, we have for all $x>x_0$ that $f'(t)-g'(t)$ is a non-negative function on $(x_0,x),$ whence $$f(x)-g(x)=\int_{x_0}^x\bigl(f'(t)-g'(t)\bigr)dt\ge0$$ for any such $x,$ and so $g(x)\le f(x)$ for such $x.$ By the third known condition, we have for all $x<x_0$ that $g'(t)-f'(t)$ is a non-negative function on $(x,x_0),$ so $$f(x)-g(x)=\int_{x_0}^x\bigl(f'(t)-g'(t)\bigr)dt=\int_x^{x_0}\bigl(g'(t)-f'(t)\bigr)dt\ge0$$ for any such $x,$ and so $g(x)\le f(x)$ for such $x.$ Since $g(x_0)\le f(x_0)$ by the first known condition, then we can conclude that $$g(x)\le f(x)$$ for all $x\in\Bbb R.$

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Edit: As Martin points out in the comments below, differentiability is sufficient for our purposes.

Define $h(x):=f(x)-g(x)$ for all $x\in\Bbb R.$ Proving that $g(x)\le f(x)$ for all $x\in\Bbb R$ is equivalent to proving that $h$ is a non-negative function.

By the first known condition, we have $$h(x_0)=0.$$ Observing that $h'(x)=f'(x)-g'(x)$ for all $x\in\Bbb R$--fairly straightforward to show by difference quotient--we have by the second known condition that $$h'(x)\ge 0\text{ for }x>x_0$$ and we have by the third known condition that $$h'(x)\le 0\text{ for }x<x_0.$$

Now, take any $x<x_0.$ Since $h$ is continuous on $[x,x_0]$ and differentiable on $(x,x_0),$ then by the Mean Value Theorem, there is some $c\in(x,x_0)$ such that $$h'(c)=\frac{h(x_0)-h(x)}{x_0-x}=-\frac1{x_0-x}h(x).$$ Since $c<x_0,$ then we have $$-\frac1{x_0-x}h(x)\le0,$ and since $x

On the other hand, given $x>x_0,$ there is likewise some $c\in(x_0,x)$ such that $$h'(c)=\frac{h(x)-h(x_0)}{x-x_0}=\frac1{x-x_0}h(x),$$ whence we likewise find that $h(x)\ge 0,$ and we're done!