1. Take f(0) = 0. Then, look at the sign of the first non-zero derivative at x = 0. If it's positive, then f(x) < 0 for x < 0 sufficiently close to zero, absurd. If it's negative, then f(x) > 0 for x > 0 sufficiently close to zero, absurd again. So all derivatives must be zero at x = 0. 2. Look at f(x) with x < 0. If f(x) > 0, then by Lagrange's theorem f'(t) < 0 for x < t < 0, done. Otherwise, f(x) = 0 for all x < 0. 3. Look at f(1) = 1, and assume all derivatives are non-negative. Remembering that f'(0) = 0, by Lagrange's theorem it must be f'(t) > 1 for 0 < t < x. Since f(x) is infinitely differentiable, this argument applies recursively. Consequently, we have: a. t_n -> 0, given by recursively applyling Lagrange's theorem, b. df / dx^n f(t_n) > 1 Therefore, lim n->+oo df/dx^n f(t_n) > 0, absurd because by now all derivatives must be zero at x = 0. Consequently, some derivative must be negative somewhere. Andres. On 12/31/18 18:45 , Andres Valloud wrote:
Hi, did you make progress on this? I ran into the same example you mentioned.
On 12/14/18 16:01 , Gareth McCaughan wrote:
On 14/12/2018 16:11, Victor Miller wrote:
Here's the essential idea: since f(0) = 0 and f(1)=1, if we define y = inf { x>= 0: f(x) > 0}, we have y in [0,1). Since f is infinitely differentiable, it can't hold that f(x) = 0 for all x <= y. So f must be decreasing somewhere before y, and thus its derivative is negative.
At the risk of once again showing myself to be a moron, that doesn't sound right. Suppose g(x) = 0 for x<=0 and g(x) = exp(-1/x^2) for x>0. Then g is infinitely differentiable but has the property you say f can't have on account of being infinitely differentiable.
(I am not claiming that g or anything like it is a counterexample to the theorem the question is asking for a proof of. It's easy to find places where, say, its second derivative is zero. But it's a counterexample to the I-think-not-a-theorem that says that an infinitely differentiable function can't be 0 for all x <= some x0 and strictly increasing thereafter.)
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