Let $f(x)$ be a nice/$C^{\infty}$ function of some real-variable $x$ which satisfies the inequality $$ |f'(x)| \leq 1\ \ \ \ \ \ \ \mathrm{for\ all\ }x>0 $$ If this is true, does this imply that $$ |f(x)| \leq |x| \ \ \ \ \ \ \ \mathrm{for\ all\ }x>0? $$ Or something similar? This seems to follow by ``integrating both sides''. I've come up with some functions where this seems to be the case (for example $f'(x) = \tanh(x)$ satisfies $|f'(x)| \leq 1$, and $f(x) = \log(\cosh(x))$ satisfies $|f(x)| < |x|$).

If this is true, how can you prove it?

If this is not true, what is an example of a counter-example?

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    $\begingroup$ What about $f(x) = 123$? $\endgroup$ – Martin R Sep 6 at 19:33
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    $\begingroup$ If $f(0)=0$ then yes, from MVT. $\endgroup$ – rtybase Sep 6 at 19:36

To elaborate a bit: the condition $$ |f'(x)| \leq 1\quad\text{ for all } x>0 $$ does not imply any bound on $f$ because $f$ can be shifted by a constant and the derivative does not change. For example, we could shift $f$ up by $100$, down by $1000$, and so on, and the derivative would not change.

A concrete counterexample is given in the comments: if $f(x) = 1000$, then $|f'(x)| = 0$ so it is bounded by $1$, but $|f(x)|$ is not bounded by $|x|$.

So, if you want the bound on $|f(x)|$, you need to additionally specify something like $f(0) = 0$. In general, the best bound you can have is $$ |f(x)| \le |x| + |f(0)|, $$ which follows by Traingle inequality from Martin's answer.

P.S.: A particular consequence of the bound on the derivative $|f'(x)| \le 1$ that your function $f$ is Lipschitz continuous with constant $1$.

  • $\begingroup$ Assuming we have the bound $|f(x) - f(0)| \leq |x|$. The triangle inequality seems to tell me that $|f(x) - f(0)| \leq |f(x)| + |f(0)|$. But I don't see how your inequality $|f(x)| \leq |x| + |f(0)|$ follows from this. What am I missing? $\endgroup$ – Greg.Paul Sep 8 at 2:08
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    $\begingroup$ @Greg.Paul There is a second useful form of triangle inequality: $|a - b| \ge |a| - |b|$. To see why this is the same as the usual triangle inequality, you can write $a = b + a'$, and the formula becomes $|a'| \ge |b + a'| - |b|$, which rearranges to $|b + a'| \le |a'| + |b|$. (Follow the steps backwards to go from the regular triangle inequality to this second form) $\endgroup$ – 6005 Sep 9 at 19:09
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    $\begingroup$ Anyway, using that form of triangle inequality, we have $|f(x) - f(0)| \ge |f(x)| - |f(0)|$. Therefore, $|f(x) \le |f(0)| + |f(x) - f(0)| \le |f(0)| + |x|$. $\endgroup$ – 6005 Sep 9 at 19:11
  • $\begingroup$ Ah I see, thanks I blanked on that one! $\endgroup$ – Greg.Paul Sep 9 at 19:56

$|f'(x)| \leq 1 $ for $x > 0$ implies that $|f(x) - f(0)| \le |x|$ for $x \ge 0$, that is an immediate consequence of the mean-value theorem.

You don't need $C^\infty$ for this conclusion, it suffices that $f$ is continuous for $x \ge 0$ and differentiable for $x > 0$.

If in addition $f(0) = 0$ then indeed $|f(x)| \le |x|$.

  • $\begingroup$ The MVT says that $f$ needs to be continuous on $[a,b]$ and differentiable on $(a,b)$ and then there always exists a $a<c<b$ where $f'(c) = [ f(b) - f(a) ] / (b-a)$. Here is seems like you're picking $a=0$ and $b=x$, and then $c=x$. This doesn't seem kosher to me because $b=c$ here...what am I misunderstanding? $\endgroup$ – Greg.Paul Sep 8 at 1:54
  • $\begingroup$ Ahh, never mind. I got it now. Pick $a=0$ and $b=x > 0$. By the MVT there exists a $0 < c < x$ where $f'(c) = \frac{f(x) - f(0)}{x}$. The bound $|f'(c)| \leq 1$ then implies that $| \frac{f(x) - f(0)}{x} | \leq 1$ from which the statement follows. Thanks! $\endgroup$ – Greg.Paul Sep 8 at 2:05

Take $$f(x)=\cos(x)$$


$f$ is $C^{\infty}$ and $$|f'(x)|\le 1$$




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