Let U ⊆ R be open, f : U → R a convex function and m t ( x ) := f ( x + t ) + f ( x − t ) − 2 f ( x ) 2 .We may suppose that U = ( a , b ) wehre a ∈ R ∪ { − ∞ } and b ∈ R ∪ { + ∞ }. Note that convexity of f implies that m t ( x ) is non-negative.Further, let P ( A ) := ∫ A e − f ( x ) d x be a probability measure.Denote N t := { x ∈ R : m t ( x ) = 0 } and suppose that P ( N t ) &gt; 1 4 for all t &gt; 0.Is it true (and is there a simple way to prove) that f is constant on U?It seems to me that this should hold but I have not been able to prove it yet.Thank you.

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Let   U  ⊆      R   be open,   f  :  U  →      R   a convex function and       m    t    (  x  )  :=            f      (      x      +      t      )      +      f      (      x      −      t      )      −      2      f      (      x      )        2  .We may suppose that   U  =  (  a  ,  b  ) wehre   a  ∈      R    ∪  {  −  ∞  } and   b  ∈      R    ∪  {  +  ∞  }. Note that convexity of   f implies that       m    t    (  x  ) is non-negative.Further, let       P    (  A  )  :=      ∫    A        e          −      f      (      x      )        d  x be a probability measure.Denote       N    t    :=  {  x  ∈      R    :      m    t    (  x  )  =  0  } and suppose that       P    (      N    t    )  &amp;gt;      1    4   for all   t  &amp;gt;  0.Is it true (and is there a simple way to prove) that   f is constant on   U?It seems to me that this should hold but I have not been able to prove it yet.Thank you.

Bornejecbo · Accepted Answer

Counterexample. Let   f  (  x  )  =      |    x      |   on   U  =      R  . Then       m    t    (  x  )  =  0 whenever       |    x      |    &amp;gt;      |    t      |  . In particular,   λ  (      N    t    )  =  ∞.Now to answer the new question. I will do the case   U  =      R  , otherwise you have to think about that mt means if   x  +  t or   x  −  t are not in   U.Since we have a probability measure, there exists   c  &amp;lt;  d such that   P  (      R    ∖  [  c  ,  d  ]  )  &amp;lt;  1      /    4. Therefore, for every   t  &amp;gt;  0, there exists a number   x  ∈  [  c  ,  d  ] such that       m    t    (  x  )  =  0. Since then   f is linear on   [  x  −  t  ,  x  +  t  ], by taking   t arbitrarily large, we obtain that   f is linear on       R  . (And being linear,   P isn&#039;t a probability measure, so   f doesn&#039;t exist.)

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