Can a real-valued, lower-semicontinuous, convex function $f$ defined on a non-empty, closed, convex subset $S$ of a real vector space $V$, with $\mathbf{0}\in S$, be extended to a real-valued, lower-semicontinuous, convex function over $V$? If it helps, it can be assumed that $V$ is a Banach space.


The rationale for this question is the following. In a certain problem I am working on, I have

  1. a real Banach space, $\mathbf{B}$,
  2. a closed, convex subset, $\mathbf{S}$, of $\mathbf{B}$, with $\mathbf{0}\in\mathbf{S}$,
  3. a function $v : \mathbf{S}\rightarrow\mathbb{R}_+$ that is convex and lower-semicontinuous.

I would like to conclude, based on Proposition 3.1 on p. 14 of Ekeland and Temam's Convex analysis and variational problems (SIAM 1987), that $v$ is the supremum of all the affine functions that are less-than-or-equal to $v$. However, this proposition applies to a function $F$ that is defined over a (locally convex, topological) vector space, whereas my $v$ is defined over a mere closed, convex subset of such a vector space. So one possible strategy would be to extend $v$ to the entire $\mathbf{B}$, and then apply the above proposition.

  • 1
    $\begingroup$ If you just want to apply the result from E/T then extending $v$ with $+\infty$ outside $S$ should do the trick. $\endgroup$
    – daw
    May 6, 2018 at 18:29

1 Answer 1


No, this is not possible. Take $S = [-1,1]$ and $f(x) = -\sqrt{1 - x^2}$.

The problem you describe under "rationale" might be solvable, but I do not know an easy argument.

  • $\begingroup$ Your $f$ plus one works for that too $\endgroup$
    – Red shoes
    Mar 18, 2019 at 17:31

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