# $\left|f'(x)\right| \leqslant g'(x)$ implies $\left|f(x)-f(a)\right| \leqslant \left|g(x)-g(a)\right|$

Suppose that both $f$ and $g$ are real, differentiable functions over $[a, +\infty)$, where $a$ is a real number; and that for all $x \geqslant a$, $\left|f'(x)\right| \leqslant g'(x)$.

Prove that $\left|f(x)-f(a)\right| \leqslant \left|g(x)-g(a)\right|$ holds for all $x \geqslant a$.

Question 1 (this is the "main question")

This is my thought: since that the derivative measures the rate of change, from $\left|f'(x)\right| \leqslant g'(x)$ we know that $f$ changes slower that $g$, so the prop. is intuitively true: $g$ changes more than $f$ in a fixed time, i.e. $x-a$. But I cannot go any further to formalize this thought, any hint?

Question 2

Can this prop. be proved using Lagrange's Mean Value Theorem?

Question 3

I know it can be proved using some helper function, like $e^{g(x)-f(x)}$ and $e^{g(x)+f(x)}$, simply take derivative of the helper function and use the property of monotone increase of $e^x$.

I came across this method, (plugging in a $e^x$ somewhere), a lot, what's the motivation behinds this method, how did they discovered this method, or it's just a trick (if there is such thing as pure trick)?

Thanks.

• Although $|g(x) - g(a)|$ is not an error, $g(x) - g(a)$ would make more sense. – Dave Sep 6 '14 at 3:19
• @ Dave: $g$ increases so $|g(x)-g(a)|=g(x)-g(a)$. – Kim Jong Un Sep 6 '14 at 3:26
• Yes, you're right. But if you don't notice that, then the statement without the absolute value appears stronger, and implies the fact you mentioned. – Dave Sep 6 '14 at 3:32

You are correct to consider auxiliary functions.

Look at the functions $h_1(x) = g(x) - f(x)$ and $h_2(x) = g(x) + f(x)$. It will be easy for you to prove that their derivatives are nonnegative. Therefore the functions $h_1$ and $h_2$ are increasing (i.e., nondecreasing). Thus $h_1(x) \geq h_1(a)$ and $h_2(x) \geq h_2(a)$. The conclusion follows from this.

This fits with your intuition. The fact that $f'(x) \leq g'(x)$ translates into the fact that $g$ grows faster than $f$, i.e., $g-f$ increases.

I don't immediately see how to use the mean value theorem if you want to apply it directly to $f$ and $g$. It seems that it would have to be applied to $g - f$ and $g+f$. In that case, it will likely be exactly the same use of the mean value theorem that is used to prove the theorem that if $h'(x) \geq 0$ then $h(x)$ is increasing.

• Hmm, seems the $e^x$ stuff is unnecessary. – Not an ID Sep 6 '14 at 3:23
• It doesn't seem relevant. – Dave Sep 6 '14 at 3:25

One approach is via the fundamental theorem of calculus: $$|f(x)-f(a)|=\left|\int_a^x f'(s)ds\right|\leq\int_a^x|f'(s)|ds\leq\int_a^xg'(s)ds=g(x)-g(a)\leq|g(x)-g(a)|.$$ Edit: as noted in Dave's comment, this argument assumes the integrability of $f'$ and $g'$. Dave's own answer in this thread doesn't require this assumption.

• Are you certain this is correct in the event that $f'(x)$ is not continuous (or even not integrable)? – Dave Sep 6 '14 at 3:16