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I hear that $f(x)=x^4$ is a strictly convex function $\forall x \in \Re$. However, strict convexity condition is that the second derivative should be positive $\forall x \in \Re$. For the mentioned function, the second derivative is zero at $x=0$, which is in the domain of $f$. Therefore, it should not be a strict convex function. But I am pretty sure it is because I heard it in Prof. Boyd's lecture.

Am I missing something obvious?

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    $\begingroup$ If the second derivative is positive, then the function is strictly convex. The converse is not true. This is an example. $\endgroup$ – Julien Mar 5 '13 at 4:36
  • $\begingroup$ Some people call a function uniformly convex if it satisfies $f''\ge c$ for some positive constant $c$. This is a stronger property than strict convexity. $\endgroup$ – user53153 Mar 5 '13 at 4:47
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    $\begingroup$ It's analogous to $x^3$: that function is an increasing function, even though it hesitates briefly at $x=0$. $\endgroup$ – André Nicolas Mar 5 '13 at 4:53
  • $\begingroup$ @5pm: Of course, uniform convexity is stronger yet than having a positive second derivative. (Consider the basic exponential function, for example.) $\endgroup$ – Cameron Buie Mar 5 '13 at 5:00
  • $\begingroup$ Note that the second derivative is positive almost everywhere. I'm pretty sure that this is also enough to get strict convexity. $\endgroup$ – Albert Dec 14 '13 at 9:54
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If the second derivative is strictly positive, then the function is strictly convex. However, the converse need not be true. A function $f:\Bbb R\to\Bbb R$ is strictly convex if and only if for all $x,y\in\Bbb R$ with $x\neq y$ we have $$f\bigl(tx+(1-t)y\bigr)<tf(x)+(1-t)f(y)$$ for all $0<t<1$. $f(x)=x^4$ is indeed strictly convex.

This is a common misconception. Many make the same mistake regarding the relationship between positive first derivative and increasing functions (the former implies the latter, but not vice versa). See here and here, for examples of people making such errors.

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  • $\begingroup$ Ok. So the converse is not true... that is even if second derivative at x=0 is zero, we say that $x^4$ is strictly convex. Now my question is how to analytically see if the function is indeed strictly convex? (In case, we cannot plot the function and see if the line segment joining any two points lies above the function) $\endgroup$ – Parag S. Chandakkar Mar 5 '13 at 5:00
  • $\begingroup$ So is there any way to prove an arbitrary function (even or odd) is strictly convex? $\endgroup$ – Parag S. Chandakkar Mar 5 '13 at 5:06
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    $\begingroup$ You may find it easier to show that it is strictly midpoint convex--that is, that $$f\left(\frac{x+y}2\right)<\frac{f(x)+f(y)}2$$ whenever $x\neq y$. Midpoint convexity is equivalent to convexity for continuous functions. In general, though, you're probably going to have to do some grunt-work calculations. $\endgroup$ – Cameron Buie Mar 5 '13 at 5:09
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Strict Convexity is when $f(tx+(1-t)y)<tf(x)+(1-t)f(y)$. The fact that $f''(x)\geq 0$ implies $f$ is convex, however, it does not necessarily imply that $f$ is not strictly convex. In the case of $f(x)=x^4$, you get $f''(x)=12x^2\geq 0$ but in fact $f$ is strictly convex. Indeed, any line segement lies above the curve.

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Second derivative being positive is for strong convexity. There is a subtle difference between strict convexity and strong convexity. A strongly convex function is strictly convex but the converse need not be true.

The condition for strict convexity is strict Jensen's inequality as pointed out by Alex R.

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protected by Community May 17 '17 at 22:58

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