Is a continuous function with zero “Taylor approximation” smooth?

Question

Let $f:[0,\infty) \to [0,\infty)$ be a continuous, non-decreasing function, satisfying $f(0)=0$ and $f(x)=o(x^n)$ for any $n \ge 1$. Suppose also that $f(x)>0$ for every $x>0$.

I also know that $f$ is smooth on the open interval $(0,\infty)$.

Must $f$ be infinitely (right) differentiable at zero?

Comments:

The Taylor series of $f$ does not have to converge to $f$, e.g. in the famous example of $$ f(x) =\begin{cases}e^{-1/x^2} \text{ for } x >0 \\ 0 \text{ for } x = 0\end{cases} $$
In general, the existence of a polynomial approximation by itself does not imply differentiabiliy; in fact it does not even imply continuity (at $x>0$), as the following examples show:

$f(x)=\chi_{\mathbb Q}(x)x^n$ or even $\chi_{\mathbb Q}(x)e^{-1/x^2}$ which also satisfies $f(x)=o(x^n)$ for any $n \ge 1$.

@robjohn The argument works only for the first derivative. You cannot show that $f''(0)=0$. — Kavi Rama Murthy, May 05 '20 at 11:51
What do you mean by smooth on $(0,\infty)$? Do you mean infinitely differentiable? — robjohn, May 05 '20 at 11:58
By the way, your examples $\chi_{\mathbb{Q}}(x)x^n$ and $\chi_{\mathbb{Q}}(x)e^{-1/x^2}$ are continuous in $0$. — Klaus, May 05 '20 at 12:02
@robjohn Sure, but they are in fact continuous (and differentiable for suitable $n$) at $0$. — Klaus, May 05 '20 at 12:06
How can a function be twice differentiable at $0$ when its first derivative is not defined in any neighborhood of $0$? Then there is the fact that the function is not monotonic. — robjohn, May 05 '20 at 12:42

robjohn · Answer 1 · 2020-05-07T12:59:28.697

Define the family of $C^\infty$ functions $$ \varphi_n(x)=\left\{\begin{array}{cl} 0&\text{for }x\le0\\ \frac{e^{\frac{n}{1-x}}}{e^{\frac{n}{x}}+e^{\frac{n}{1-x}}}&\text{for }0\lt x\lt1\\ 1&\text{for }x\ge1 \end{array}\right. $$ then $$ \varphi_n^\prime\!\left(\frac12\right)=2n $$

With $f(x)=e^{-\frac1x}$, and $x_k=\frac1k$, define $$ \begin{align} g(x) &=f(x_k)+(f(x_{k-1})-f(x_k))\,\varphi_{n_k}\!\!\left(\frac{x-x_k}{x_{k-1}-x_k}\right)&&\text{for }x_k\le x\le x_{k-1}\\ &=e^{-k}\left(1+(e-1)\varphi_{n_k}((k-1)(kx-1))\right)&&\text{for }\frac1k\le x\le\frac1{k-1}\\ \end{align} $$

$g$ is infinitely differentiable, monotonically increasing, and $g(x)\le e^{1-\frac1x}$.

For integer $k\ge1$, $$ g'\!\left(\frac1k\right)=0 $$ and for integer $k\ge2$, $$ g'\!\left(\frac12\left(\frac1{k-1}+\frac1k\right)\right)=2n_kk(k-1)(e-1)e^{-k} $$ If we set $n_k=3^k$, then $g'(x)$ varies between $0$ and $2k(k-1)(e-1)\left(\frac3e\right)^k$ on $\left[\frac1k,\frac1{k-1}\right]$. Thus, $g''(0)$ does not exist.

score 1 · Accepted Answer · answered May 05 '20 at 13:51

1

No, $f$ doesn't have to be infinitely differentiable at $0$: define $$f(x) = \int_0^x e^{-1/t^2}\left[1 + \sin (e^{1/t^2}) \right]dt$$ for $x \ge 0$ and $f(x) = 0$ for $x \le 0$.

$f$ satisfies the hypothesis, but $\lim_{x \to 0}f''(x)$ does not exist.

answered May 05 '20 at 13:51

Sam

1,401

However, $f''(0)=0$, but $f'''(0)$ does not exist. – robjohn May 05 '20 at 20:12

Is a continuous function with zero “Taylor approximation” smooth?

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