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mynameisfunk
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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].
I see that c=\frac{1}{2}((c+k)+(c-k)). I also see that the inequality is exactly the one Rudin uses to prove that the derivative of a local maximum is 0. I don't really understand what the hint is. There is no supremum, right?
What I tried to do and decided I couldn't make it work was to take \delta > 0 and take x_0, x_1 such that d(x_1,1)=d(x_0,0)<\delta and let k=d(x_1,1)=(x_0,0) so that now g(c)=0 when c=\frac{1}{2} and I was going to show that g(c+k),g(c-k) would always have to equal 0, but then I was thinking that what if the function oscillated and intersected the x-axis at 0,1/2, and 1 so that g(c-k)=-g(c+k). Seems like it would hold for my proof, also I didnt use the hint. HELP! This problem seems easy but I can't seem to wrap my head around it.
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].
I see that c=\frac{1}{2}((c+k)+(c-k)). I also see that the inequality is exactly the one Rudin uses to prove that the derivative of a local maximum is 0. I don't really understand what the hint is. There is no supremum, right?
What I tried to do and decided I couldn't make it work was to take \delta > 0 and take x_0, x_1 such that d(x_1,1)=d(x_0,0)<\delta and let k=d(x_1,1)=(x_0,0) so that now g(c)=0 when c=\frac{1}{2} and I was going to show that g(c+k),g(c-k) would always have to equal 0, but then I was thinking that what if the function oscillated and intersected the x-axis at 0,1/2, and 1 so that g(c-k)=-g(c+k). Seems like it would hold for my proof, also I didnt use the hint. HELP! This problem seems easy but I can't seem to wrap my head around it.
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