Let $f:[a,b] \to \mathbb{R}$ be a continous, convex function. (By convex, I mean $f(\lambda x+(1- \lambda)y) \leq \lambda f(x)+(1- \lambda)f(y)$ for any choice of $x,y \in [a,b]$ and $\lambda \in [0,1]).$

Q: Is $f$ Lipschitz? If not, what would be a counterexample?

I know a theorem that says a convex function on $(a,b)$ has to be Lipschitz on each $[c,d]$. However, with no extra assumptions, the Lipschitz constant might change.

Will continuity ensure that there's one Lipschitz constant that works for everything?

  • $\begingroup$ square root of $x$ on $[0,1]$ is not a counter-example? $\endgroup$
    – Leandro
    Feb 26, 2020 at 19:35
  • 2
    $\begingroup$ $\sqrt x $ is concave, but you might be on to something :) $\endgroup$ Feb 26, 2020 at 19:36
  • $\begingroup$ $x \mapsto -\sqrt{x}$ is convex on $[0,1]$. $\endgroup$
    – copper.hat
    Feb 26, 2020 at 19:47
  • 1
    $\begingroup$ $x^x$ is also interesting. $\endgroup$
    – zwim
    Feb 26, 2020 at 20:12

2 Answers 2


As a convex function, $f$ has a left and right derivative in every point of $(a, b)$, and these are monotonically increasing. But these (one-sided) derivatives can approach $-\infty$ for $x \to a$ or $+\infty$ for $x \to b$, and then $f$ is not Lipschitz continuous on $[a, b]$.

An example is $f(x) = 1 - \sqrt{x}$ on $[0, 1]$. It is convex, but the derivative approaches $-\infty$ for $x \to 0$, so that it is not Lipschitz continuous.

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On the unit circle, look at the arc from $(0,-1)$ to $(1,0).$ That is the graph of a convex function that has derivative $+\infty$ at $(1,0).$ Right there you have a counterexample.


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