§ Flat functions

Define
f(x)≡{0x<=0e−1/xx>0 f(x) \equiv \begin{cases} 0 & x <= 0 \\ e^{-1/x} & x > 0 \end{cases}
This is smooth, but is badly non-analytic. Any taylor expansion around x=0x=0 is going to be identically zero. So we're going to prove that it possesses all derivatives. This implies that the derivative at zero is equal to zero, because the left derivative is always equal to zero.
  • f(x)≡e−1/xf(x) \equiv e^{-1/x}. Differentiate to get f′(x)=e−1/x/x2f'(x) = e^{-1/x}/x^2. Change y≡1/xy \equiv 1/x to get y2e−yy^2 e^{-y}. As y↦∞y \mapsto \infty, e−ye^{-y}decays more rapidly than y2y^2 increases, thus the limit is zero. Hence, f′(x)=0f'(x) = 0.
  • For higher derivatives, let f(n)(x)≡pn(1/x)e−1/xf^{(n)}(x) \equiv p_n(1/x) e^{-1/x} for some polynomial pnp_n. See that f(n+1)(x)=d/dx[pn(1/x)e−1/x)]f^{(n+1)}(x) = d/dx [p_n(1/x) e^{-1/x})]. To compute this, set y≡1/xy \equiv 1/xand compute d/dy[pn(y)e−y]dy/dxd/dy [p_n(y) e^{-y}] dy/dx which is:
=d/dy[pn(y)e−y]dy/dx=pn′(y)e−y+pn(y)(−e−y)⋅1/x2=e−1/x(pn′(1/x)−pn(1/x))1/x2=e−1/x(qn′(1/x)−qn(1/x))let rn+1(x)(≡qn′(t)−qn(t))t2=rn+1(1/x)e−1/x \begin{aligned} & = d/dy [p_n(y) e^{-y}] dy/dx \\ & = p_n'(y) e^{-y} + p_n(y) (- e^{-y}) \cdot 1/x^2 \\ & = e^{-1/x} (p_n'(1/x) - p_n(1/x)) 1/x^2 \\ & = e^{-1/x} (q_n'(1/x) - q_n(1/x)) \\ & \text{let $r_{n+1}(x) (\equiv q_n'(t) - q_n(t))t^2$} \\ & = r_{n+1}(1/x) e^{-1/x} \end{aligned}
  • So we can write higher derivatives too as poly(1/x)poly(1/x) times exp(−1/x)exp(-1/x) which also decays rapidly to 00.
  • Philosophically, what's going on is that a non-zero polynomial can only have a finite number of zeroes. Since this function has an infinite number of zeroes around it's neighbourhood at (x=0)(x = 0), any polynomial that agrees with this function in any neighbourhood must be identically zero everywhere.