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Let $f : \mathbb{R}^N \longrightarrow \mathbb{R} \cup \{+\infty \}$ be a lower semi-continuous convex proper function.

Let $dom f$ be the domain of $f$, i.e. $dom f:= \{ x \in \mathbb{R}^N \ | \ f(x) <+\infty \}$.

I am interested in the continuity of $f$ relative to the domain, that is, in the following property:

$$x_n,x \in dom f \text{ and } x_n \to x \text{ implies that } f(x_n) \to f(x).$$

It is well-known that convex functions are continuous on the interior of their domain (even on the relative interior of their domain). For instance, consider

$$\begin{array}{c|l} f(x)= & 0 \text{ if } x \in [0,1[, \\ & 1 \text{ if } x = 1, \\ & + \infty \text{ else. } \end{array}$$ It is convex, and clearly continuous relatively to $[0,1[$, but observe that it is not lower semi-continuous there.

So my question is: can you provide a counter-example of lower semi-continuous convex proper function which would have a point of "discontinuity" on the boundary of its domain? That is, finding a converging sequence $x_n \to x$ such that both $x_n$ and $x$ lie in the domain, but such that $f(x) < \liminf f(x_n)$?

Hint 1: It seems to me that the counter example shall be found with $N \geq 2$.

Hint 2: Maybe a counter-example can be found by finding a sequence for which the sequence of values $f(x_n)$ has 2 limit points.

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Take the function $\phi:S^1\to\mathbb R$ defined by $\phi(x,y)=1$ if $(x,y)\neq(1,0)$ and $\phi(1,0)=0$. Let f be the convex envelope of $\phi$ on the disk, that is $$f(x)=\sup\{h(x)| h \textrm{ affine and }h<\phi \textrm{ on } S^1\}.$$ It can be easily shown that on the restriction of f on the chord joining $(1,0)$ to any $(x,y)$ on $S^1$ is the affine interpolation between $0=f(1,0)$ and $1=f(x,y)$. Thus f is not continuous on the closed disk.

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