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Can anyone see where the flaw is in the development below, where I prove that (g o f)'(x)=g'(f(x)) instead of g'(f(x))f'(x), as it should be.
Consider the usual hypothese under which the chain rule for real-valued function applies.
Consider \epsilon>0. Since g is differentiable at f(x_0), there exists \delta_1(\epsilon)>0 such that |y-f(x_0)|<\delta_1(\epsilon) implies |g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq\epsilon|y-f(x_0)|
On the other hand, f being differentiable at x_0, enjoys the Lipschitz property there. That is to say, there exists positive constants c and M such that |x-x_0|<c implies |f(x)-f(x_0)|\leq M|x-x_0|.
So for \delta=\min(c,\delta_1(\epsilon/M)/M), we have that |x-x_0|<\delta implies |f(x)-f(x_0)|\leq M|x-x_0|<M\delta\leq M\delta_1(\epsilon/M)/M= \delta_1(\epsilon/M)
which in turns implies that
|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq(\epsilon/M) |f(x)-f(x_0)|\leq (\epsilon/M)M|x-x_0| = \epsilon |x-x_0|
And this is equivalent to saying that (g o f)'(x) = g'(f(x))...
Consider the usual hypothese under which the chain rule for real-valued function applies.
Consider \epsilon>0. Since g is differentiable at f(x_0), there exists \delta_1(\epsilon)>0 such that |y-f(x_0)|<\delta_1(\epsilon) implies |g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq\epsilon|y-f(x_0)|
On the other hand, f being differentiable at x_0, enjoys the Lipschitz property there. That is to say, there exists positive constants c and M such that |x-x_0|<c implies |f(x)-f(x_0)|\leq M|x-x_0|.
So for \delta=\min(c,\delta_1(\epsilon/M)/M), we have that |x-x_0|<\delta implies |f(x)-f(x_0)|\leq M|x-x_0|<M\delta\leq M\delta_1(\epsilon/M)/M= \delta_1(\epsilon/M)
which in turns implies that
|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq(\epsilon/M) |f(x)-f(x_0)|\leq (\epsilon/M)M|x-x_0| = \epsilon |x-x_0|
And this is equivalent to saying that (g o f)'(x) = g'(f(x))...
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