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Let $E$ be a normed space and $f:E \to \mathbb R$ convex.

  • If $E = \mathbb R^d$ then $f$ is locally Lipschitz-continuous.
  • If $E$ is infinite-dimensional then $f$ is not necessarily continuous. There exists discontinuous linear functional on an infinite-dimensional normed space.

Now we assume more that $f$ is strictly convex, i.e., $$ f(tx + (1-t)y) < tf(x)+(1-t)f(y) \quad \forall t \in (0, 1), \forall x,y\in E \text{ s.t. } x\neq y. $$

  • Strict convexity does not imply differentiability.

Does strict convexity of $f$ imply that $f$ is continuous?

Analyst
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1 Answers1

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No. Just add any continuous, strictly convex function (e.g. the norm squared) to a discontinuous linear functional. The sum of a convex $f$ and strictly convex $g$ must be strictly convex, as \begin{align*} (f + g)(\lambda x + (1 - \lambda)y) &= f(\lambda x + (1 - \lambda)y) + g(\lambda x + (1 - \lambda)y) \\ &\le \lambda f(x) + (1 - \lambda)f(y) + g(\lambda x + (1 - \lambda)y) \\ &< \lambda f(x) + (1 - \lambda)f(y) + \lambda g(x) + (1 - \lambda)g(y) \\ &= \lambda (f + g)(x) + (1 - \lambda)(f + g)(y) \end{align*} for distinct $x, y \in E$ and $\lambda \in (0, 1)$. Also, if $f + g$ were continuous at any point, then so would be $f = (f + g) - g$, by the algebra of continuous functions.

Theo Bendit
  • 50,900