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Prove that a function has a Taylor series that converges everywhere, but does not represent the function

by RoRi

Define a function

    \[ f(x) = \begin{cases} e^{-\frac{1}{x^2}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0. \end{cases} \]

  1. Prove that f(x) has derivatives of every order everywhere on \mathbb{R}.
  2. Prove that the nth derivative at 0, f^{(n)}(0), is equal to 0 for all n \geq 1. This shows that the Taylor series for f at the point x = 0 converges everywhere on \mathbb{R} but it does not represent the function away from the origin.

We prove parts (a) and (b) at the same time. Then, we prove Apostol’s comment in part (b).

Proof. We show a bit more than requested, not only that f^{(n)}(0) exists for all n, but in fact, f^{(n)}(0) = 0 for all n \geq 1. The proof is by induction. For the case n = 1 we have,

    \begin{align*} f'(0) &= \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0} \\[9pt] &= \lim_{x \rightarrow 0} \frac{e^{-1/x^2}}{x}\\[9pt] &= \lim_{x \rightarrow 0} \frac{x^{-1}}{e^{1/x^2}}\\[9pt] &= \lim_{x \rightarrow 0} \frac{-1/x^2}{(-2/x^3) e^{1/x^2}} & (\text{L'Hopital})\\[9pt] &= \lim_{x \rightarrow 0} \frac{x}{2 e^{1/x^2}}\\[9pt] &= \lim_{x \rightarrow 0} P(x) e^{-1/x^2} & (P \text{ a poly. in } x)\\[9pt] &= 0 &(x \rightarrow 0, 2e^{1/x^2} \rightarrow +\infty). \end{align*}

Now, assume we have shown f^{(k)}(0) = \lim_{x \rightarrow 0} P_k (x) e^{-1/x^2} = 0 for some integer k. We then consider f^{(k+1)}(0),

    \begin{align*} f^{(k+1)} (0) &= \lim_{x \rightarrow 0} \frac{f^{(k)}(x) - f^{(k)}(0)}{x - 0}\\[9pt] &= \lim_{x \rightarrow 0} \frac{P_k (x) e^{-1/x^2}}{x} \\[9pt] &= \lim_{x \rightarrow 0} \frac{P_k (x)/x}{e^{1/x^2}} \\[9pt] &= \lim_{x \rightarrow 0} \frac{xP_k'(x) + 2P_k(x) / x^2}{-2/x^3 e^{1/x^2}} & (\text{L'Hopital})\\[9pt] &= \lim_{x \rightarrow 0} \frac{P_{k+1}(x)}{e^{1/x^2}} & \left(\text{multiply by } \frac{x^3}{x^3}\right)\\[9pt] & = 0. \end{align*}

Since P_{k+1}(x) \rightarrow c for any polynomial P in x, and e^{1/x^2} \rightarrow + \infty as x \rightarrow 0. Thus, we indeed have f^{(n)}(0) = 0 for all n \geq 1. \qquad \blacksquare

Proof. Since we have shown in part (a) that f^{(n)}(0) = 0 for all n, the Taylor’s series generated by f about 0 is just the zero function. This clearly converges everywhere on \mathbb{R}; however, for x \neq 0, f(x) = e^{-1/x^2} \neq 0 since it is always e^x \neq 0 for any x. Therefore, we see that the Taylor series generated by f about 0 converges everywhere on \mathbb{R} but only equals f at 0. \qquad \blacksquare

(Note: This turns out to be an important example. In real analysis the fact that we can have a Taylor expansion for a function that converges everywhere, but does not represent the function breaks a lot of theorems we wish were true. It turns out this is why complex analysis is so much nicer than real analysis. In the complex case if we have a Taylor expansion then it does represent the function. Anyway, this example will come back many times if you keep doing analysis, so it’s worth understanding it now.)

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