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Lets say f is defined on an Interval $I = [a,b] $. Since f is convex, one immediately knows that f is continuous on $I^°$ , however left are the points $a$ and $b$

The piecewise continuity of f states, that f is continuous on certain open intervals and that the one sided limit of the endpoints of those interval exist.(but don't have to be the same!)

In an exercise it was postulated, that f is then also continuous at the endpoints a,b of f.

However I don't see, how this holds, if just the limits of the endpoints exist due piecewise continuous.

I think, there is an error in the definition of piecewise continuous, as long as one doesn't know that the one sided limits of f'(a), f'(b) exist, f(a) and f(b) should not have to be necessarily continuous. I am a bit unsure about this and would like to have a 2th opinion..

As always thanks in advance.

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  • $\begingroup$ If a function is $0$ on $a$ and $b$ and $1$ on $(a,b)$, is it considered piecewise continuous? It kind of depends on the definition! The definition I know, is that the set of discontinuity points is finite. $\endgroup$
    – Zardo
    Jul 20, 2015 at 9:36
  • $\begingroup$ @Zardo your function is concave :D $\endgroup$
    – user251257
    Jul 20, 2015 at 9:46
  • $\begingroup$ @user251257 Well, it doesn't really change much! Is it considered piecewise continuous though? $\endgroup$
    – Zardo
    Jul 20, 2015 at 9:47
  • $\begingroup$ @Zardo I would. $\endgroup$
    – user251257
    Jul 20, 2015 at 9:49

1 Answer 1

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The function $f$ need not be continuous at $a$ and $b$. Consider the example $$f(x):=x^2\quad(-1<x<1),\qquad f(\pm1):=2\ .$$ This $f$ is convex on $[{-1},1]$. It should however, not be too difficult to prove that the limits $\lim_{x\to a+}f(x)$ and $\lim_{x\to b-}f(x)$ exist and are finite if $f$ is defined on all of $[a,b]$.

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  • $\begingroup$ yeah, but I am not sure if $f$ is piecewise continuous then, for example $ f'(+1), f'(-1) $ don't exist. (Interestingly I used the same example for myself to see that convex functions don't necessarily have to be convex at the end points!) $\endgroup$
    – Imago
    Jul 20, 2015 at 9:59
  • $\begingroup$ @Imago why does derivative matter for piecewise continuous? that would be a strange definition $\endgroup$
    – user251257
    Jul 20, 2015 at 10:08
  • $\begingroup$ 1. I have seen this definition a couple of times 2. f is supposed to be continuous at the end points 3. one sided derivatives would instantly get me the continuity $\endgroup$
    – Imago
    Jul 20, 2015 at 10:17
  • $\begingroup$ @Imago differentiable implies continuous in general. the converse is true in case of convex functions at the end points. so it is like you directly demand continuity at the end points. $\endgroup$
    – user251257
    Jul 20, 2015 at 10:24
  • $\begingroup$ ok, ty for the clarification :) $\endgroup$
    – Imago
    Jul 20, 2015 at 10:28

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