suppose that g:[0,1] \rightarrow \re is continuous, g(0)=g(1)=0 and for every c \in (0,1), there is a k > 0 such that 0 < c-k < c < c+k < 1 and g(c)=\frac(1}{2} (g(c+k)+g(c-k)).
Prove that g(x) = 0 for all x \in [0,1] Hint: Consider sup{x \in [0,1] | f(x)=M } where M is maximum of f on [0,1]...