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All the answers (example) I've seen refer to a constant one could add to $f$ or $g$ such that the strict implication does not hold.

However, if we put a very simple constraint: $f(0) = g(0) = 0$ (e.g. there are no constants), does the implication hold (over some set of real numbers $I \subseteq R$):

$\forall x \in I.f(x) \leq g(x) \implies \forall x \in I. f'(x) \leq g'(x)$

Added after the question was already answered: Even though the answers below show that the constraint $f(0) = g(0) = 0$ is not enough, please do add the answer if there is a non-trivial constraint on $f$ and $g$ (e.g. like that $f'(x) \le g'(x)$) which would make the implication to hold.

S11n
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2 Answers2

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I don't believe there is a non-trivial additional assumption under which the implication holds. In a sense, the relation between function values doesn't say anything about their derivatives. This is illustrated by the following example:

$$\begin{align*} f(x) & = \sin^2 x \\ g(x) & = 2 \arctan^2 x \end{align*}$$

Graphs

As you can see on the graph, $\sin^2 x \leqslant 2 \arctan^2 x$ for all $x \in \mathbb{R}$. But the derivative of $\sin^2 x$ goes up and down periodically while the derivative of $2 \arctan^2 x$ tends to zero for large enough $x$, so there is no consistent relation between them.

Adayah
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A counterexample would be any pair of functions $f$ and $g$ such that:

  • $f(0)=g(0)=0$
  • There is an $a$ such that $f(a)\le g(a)$ and $f'(a)>g'(a)$.

For simplicitly, I will assume $f$ and $g$ are differentiable functions defined on $\mathbb R$.

If we let $h=g-f$, then finding a counterexample boils down to finding a function $h$ such that $h(0)=0$, and for which there is an $a$ satisfying $h(a)\ge0$ and $h'(a)<0$. This is pretty easy: for instance, we could take $h(x)=-x$. Now let $g$ and $f$ be any two differentiable functions satisfying $g(x)-f(x)=-x$ for all $x\in\mathbb R$. For instance, we could set $g(x)=\sin x$ and $f(x)=x+\sin x$.

Joe
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    You are interpreting the condition as $(\forall x \in \mathbb{R}) \big( f(x) \leqslant g(x) \implies f'(x) \leqslant g'(x) \big)$. While the original condition is not precise and does need an interpretation, I think it is slightly more probable that $(\forall x \in \mathbb{R}) , f(x) \leqslant g(x) \implies (\forall x \in \mathbb{R}) , f'(x) \leqslant g'(x)$ is what the asker had in mind. – Adayah Sep 16 '22 at 10:38
  • Thank you @Adayah for clarification and answer above, but I actually thought of any open interval over which the left side of the implication holds, that over the same interval the right hand side of the implication should hold. I clarified the question. – S11n Sep 16 '22 at 13:29