5
$\begingroup$

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $\left[0,1\right]$. Consider $z\in\left(0,1\right)$ and suppose that $f$ is differentiable and convex on $\left[0,z\right]$ and $\left[z,1\right]$. If $f'$ (i.e., $\frac{df}{dx}$) is continuous at $z$, then $f$ is convex on $\left[0,1\right]$. - Is this proposition true? While I could not create a counter-example, I am finding it difficult to generate a clean proof as well.

$\endgroup$
  • $\begingroup$ OK, I got your answers, and thank all of you for that. However, I would like to add something more to this question. To what extent, can I relax continuity of $f'$, and still have global convexity? Following your lines of proof, it seems to me that as long as the left-side derivative ($f'_{-}$) at $z$, which is well defined, is less than the right-side derivative ($f'_{+}$) at $z$, which is well defined as well, global convexity prevails. I do not need continuity of $f'$ precisely - Is this correct? $\endgroup$ – Tapas Aug 14 at 15:26
3
$\begingroup$

A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

$\endgroup$
  • $\begingroup$ Thanks for your answer. If I relax continuity of $f'$ at $z$, and just assume that the left-side derivative of $f$ at $z$ is less than the right-side derivative of $f$ at $z$, global convexity will still hold - is it right? In fact, I was wondering that is an "if and only if" condition to maintain global convexity. $\endgroup$ – Tapas Aug 14 at 15:42
  • 1
    $\begingroup$ Yes, you are right it generalizes in the way you suggest it. But note that this does not follow from the proof I gave because the criterion I used needs that $f$ is differentiable in the whole interval. If you weaken this assumption you will have to proof it in a different way. I think it will boil down to using the definition of convexity and show it explicitly like one would show the theorem I used in my answer. $\endgroup$ – Stefan Egger Aug 14 at 16:16
3
$\begingroup$

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that $$f'(x)\leq f'(z)\leq f'(y)$$ for any $0\leq x<z<y\leq 1$. Can you take it from here and show that the proposition is true?

$\endgroup$
1
$\begingroup$

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.