Let $\mathcal{H}$ denote am infinite-dimensional Hilbert space of functions and let $L(f)$ denote a convex, continuous functional over $\mathcal{H}$. I would like to know under which conditions the optimization problem

$$\min L(f) \ s.t. ||f||_{\mathcal{H}} = 1, $$

has a solution in the closed and bounded set $||f||_{\mathcal{H}} = 1$. Obviously the unit ball is not compact in infinite-dimensional spaces. Could someone point me towards a related existence theorem? Thank you in advance.


1 Answer 1


Solutions do not exist in general under these assumptions: Let $H=l^2$, $L$ defined by $$ L(f) = \sum_{k=1}^\infty k^{-1} f_k^2, $$ where $f = (f_k)$. This functional is convex and continuous, the infimum of $L$ on the unit sphere is $0$ (take $f$ equal to the unit vectors), but the infimum is not attained.

In order to get existence, some additional compactness needs to be present: $L$ having compact level sets, $L$ being completely continuous, etc.

  • $\begingroup$ Thanks a lot for your answer. Do you have maybe a reference for those existence theorems? $\endgroup$
    – JohnK
    Commented Feb 10, 2020 at 15:10
  • $\begingroup$ Can this be salvaged if one assumes that L is weakly lower semi-continuous? I would think that by the boundedness of $\{f, |f||=1\}$ and a subsequence argument, one can show that the infimum is attained. $\endgroup$
    – JohnK
    Commented Feb 12, 2020 at 12:26
  • 1
    $\begingroup$ This is not sufficient, you would also need weak (sequential) closedness of the feasible set. The sphere is not weakly closed. $\endgroup$
    – daw
    Commented Feb 12, 2020 at 12:55
  • $\begingroup$ Ah yes, indeed. Thanks a lot. $\endgroup$
    – JohnK
    Commented Feb 12, 2020 at 12:58

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