It is easy to see that f(-1)=f(1) = 0 Make the substitution: x = cos(y), then the equation becomes: f(cos(2y)) = 2 cos (y) f( cos(y)), y in [0 pi] Let g(y) = f(cosy), so we have g(2y) = 2 cos(y) g(y) -- call this equation 1 hence g(2(pi-y)) = 2 cos(pi-y) g(pi-y) = -2 cos(y) g(pi-y) Since g(2y) = g(2(pi-y)), we have so 2cos(y)g(y) = -2 cos(y) g(pi-y), hence g(pi-y) = - g(y) for any y != pi/2. By continuity, g(pi/2) = - g(pi/2), hence g(pi/2) = 0, hence for any y in [0 pi], we have g(pi-y) = -g(y) -- call this equation 2 Now, assume the conclusion does not hold, then the set A = { y|g(y)!=0 } must be non-empy and open, so it can be represented as union of non overlapping intervals. We can pick one of intervals with the largest length. Since g(pi/2) = 0, this interval is either a subset of [0 pi/2] or [pi/2 pi]. By equation 1, it can not be a subset of [0 pi/2] By equation 1 and 2, it can not be a subset of [pi/2 pi] Contradiction. |
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