Subtracting a constant from log-concave function preserves log-concavity, if the difference is positiveI am trying to work out a question from 'Convex Optimization - Boyd' . Specifically, exercise 3.48:Show that if f : R n → R is log-concave and a &gt; 0, then the function g = f − a is log-concave, where dom ⁡ g = { x ∈ dom ⁡ f | f ( x ) &gt; a } ?This is my attempt so far:By assumption, we have d 2 log ⁡ f d x 2 = f ″ f − ( f ′ ) 2 f 2 ≤ 0.Considering g = f − a, we have d 2 log ⁡ g d x 2 = f ″ ( f − a ) − ( f ′ ) 2 ( f − a ) 2 = f ″ f − ( f ′ ) 2 ( f − a ) 2 − f ″ a ( f − a ) 2 .If f ″ &gt; 0, then we are complete. I'm not sure how to handle the alternative situation though, can anyone help?

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Subtracting a constant from log-concave function preserves log-concavity, if the difference is positiveI am trying to work out a question from &#039;Convex Optimization - Boyd&#039; . Specifically, exercise 3.48:Show that if   f  :            R        n    →      R   is log-concave and   a  &amp;gt;  0, then the function   g  =  f  −  a is log-concave, where   dom  ⁡  g  =  {  x  ∈  dom  ⁡  f      |    f  (  x  )  &amp;gt;  a  }  ?This is my attempt so far:By assumption, we have                    d        2            log      ⁡      f              d              x        2              =                    f        ″            f      −      (              f        ′                    )        2                    f      2        ≤  0.Considering   g  =  f  −  a, we have                    d        2            log      ⁡      g              d              x        2              =                    f        ″            (      f      −      a      )      −      (              f        ′                    )        2                    (      f      −      a              )        2              =                    f        ″            f      −      (              f        ′                    )        2                    (      f      −      a              )        2              −                    f        ″            a              (      f      −      a              )        2              .If       f    ″    &amp;gt;  0, then we are complete. I&#039;m not sure how to handle the alternative situation though, can anyone help?

Edward Chase · Accepted Answer

Hint:You correctly compute                    d        2            log      ⁡      g              d              x        2              =                    f        ″            (      f      −      a      )      −      (              f        ′                    )        2                    (      f      −      a              )        2              .When   a  =  0, we know this quantity is non-positive, so the numerator is non-positive. Focus just on the numerator. What happens to the numerator when   a increases from   0? Can you say whether this expression remains non-positive? Note that by assumption,   f  −  a  &amp;gt;  0. It may be useful to consider the cases when       f    ″    &amp;gt;  0 and       f    ″    ≤  0 separately.And by the way, a complete proof would take account of the fact that   f is a function of values in             R        n  , not just       R  . That&#039;s really not a big issue here, though, since   f is log-concave if for all       x    0    ,  v  ∈            R        n  , the function             f      ~        :      R    →      R   is log-concave, where            f      ~        (  t  )  ≡  f  (      x    0    +  v  t  )  .This is just a simple trick for converting questions of concavity in   n dimensions into equivalent questions in one dimension by considering how   f changes along arbitrary directions   v.

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