Do lim x ↓ 0 f ( x ) and lim x ↑ 1 f ( x ) exist if f is concave over [ 0 , 1 ]?

Question

Do        lim          x      ↓      0        f  (  x  ) and        lim          x      ↑      1        f  (  x  )  exist if    f  is concave over    [  0  ,  1  ]?

Arely Davila · Accepted Answer

If   f is concave on   [  0  ,  1  ] then                    (*)                    f        (        b        )        ≥                              c            −            b                                c            −            a                          f        (        a        )        +                              b            −            a                                c            −            a                          f        (        c        )            for   0  ≤  a  &amp;lt;  b  &amp;lt;  c  ≤  1. This can be rewritten as            c      −      b              c      −      a                  (        f  (  b  )  −  f  (  a  )            )        ≥            b      −      a              c      −      a                  (        f  (  c  )  −  f  (  b  )            )          .In particular we have the following implication:                    (**)                                          (                          0        ≤        a        &amp;lt;        b        &amp;lt;        c        ≤        1         and         f        (        a        )        &amp;gt;        f        (        b        )                              )                                  ⟹                f        (        b        )        &amp;gt;        f        (        c        )                .            Now consider two cases:  f is increasing on   (  0  ,  1  ).   (  ∗  ) with   0  =  a  &amp;lt;  b  =  1      /    2  &amp;lt;  c  &amp;lt;  1 gives an upper bound for   f  (  c  ), so that       lim          x      →      1      −        f  (  x  ) exists as a finite value.Otherwise there is   0  &amp;lt;      x    1    &amp;lt;      x    2    &amp;lt;  1 with   f  (      x    1    )  &amp;gt;  f  (      x    2    ). Then   (  ∗  ∗  ) implies that f is decreasing on   [      x    2    ,  1  ).   (  ∗  ) with   a  =  0  &amp;lt;  b  &amp;lt;  c  =  1 gives a lower bound for   f  (  b  ), so that       lim          x      →      1      −        f  (  x  ) exists as a finite value in this case as wellA similar argument works for       lim          x      →      0      +        f  (  x  ).It follows from the concavity condition   (  ∗  ) that      lim          x      →      0      +        f  (  x  )  ≥  f  (  0  )        lim          x      →      1      −        f  (  x  )  ≥  f  (  1  )

Do lim_(x downarrow 0) f(x) and lim_(x uparrow 1) f(x) exist if f is concave over [0,1]?

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