# if $f(x)$ is even and can be infinitely differentiable, how about $f(\sqrt{x})$

I have a question $f(x)$ is even and can be infinitely differentiable, how about $f(\sqrt{x})$ in [0,$\infty$)? can we say that the $f(\sqrt{x})$ also can be infinitely differentiable in $[0,\infty)$.

My thoughts: let $g(x)=f(\sqrt{x})$, and then I have proved $g(x)$ can be differentiable and continuous in $[0,\infty)$, and I supposed this conclusion still holds for $n=k$($k$ is an integer),which means $g(x)$ can be $k$-times differentiable and continous in $[0,\infty)$, so I want to use mathematical induction to prove it. But I am stuck here, since I have no idea how to prove g can be $(k+1)$ times differentiable using the assumption that $g$ can be $k$-times differentiable and continuous. Can someone tell me whether it is true or not?

• this peoblem is from my professor Feb 1, 2015 at 3:38
• I think it is true, but I am struggling for it Feb 1, 2015 at 3:42
• Have you tried using the chain rule? Feb 1, 2015 at 3:51
• The proof is trivial for analytic functions. I wonder if some argument by continuity could reach the smooth case (we'd have to choose some unusual metric/topology to make such an argument work though) Feb 1, 2015 at 3:55
• $limit_{x->0} {{f(\sqrt{x})-f(0)}\over x}$ Feb 1, 2015 at 4:08

Suppose $f$ is even and infinitely differentiable, and let $g(x) = f(\sqrt{x})$. I claim that for each positive integer $k$, $g^{(k)}$ is defined on $[0, \infty)$, and moreover that $g^{(k)}(x^2)$ is an infinitely differentiable even function on $\Bbb{R}$.

Suppose that the claim holds for $k=1$, and letsuppose that $g^{(n)}(x)$ is defined for $x \ge 0$ and that $g^{(n)}(x^2)$ is even and infinitely differentiable. Then, we may define $h(x) = g^{(n)}(\sqrt{x})$, and from the base case we find that $h$ is differentiable and $h'(x^2)$ is even and infinitely differentiable. But $h(x) = g^{(n)}(x)$, so $g^{(n)}$ satisfies the required conditions. It follows that $g$ is infinitely differentiable, so long as we can prove that $g'(x)$ exists for $0 \le x$ and $g'(x^2)$ is infinitely differentiable. Let's try to prove that.

We have that for $x > 0$,

$$g'(x) = \frac{f'(\sqrt{x})}{2\sqrt{x}}$$

by the chain rule, and for $x=0$,

$$g'(0) = \lim_{h\to0} \frac{g(h) - g(0)}{h} = \lim_{h\to0} \frac{f(\sqrt{h}) - g(0)}{h}= \lim_{h\to0} \frac{f(0) + f'(0)\sqrt{h} + \frac{1}{2}f''(0)h + o(h) - f(0)}{h}$$

by Taylor's Theorem. Now, since $f$ is even, $f'(0) = 0$, so

$$g'(0) = \frac{1}{2}f''(0)$$

Define $h(x) = g'(x^2)$. Then,

$$h(x) = \begin{cases} \frac{f'(x)}{2x} & x \not= 0 \\ \frac{f''(0)}{2} & x = 0 \end{cases}$$

Since $f$ is even, we have at once that $f'(x)$ is odd, so $\frac{f'(x)}{2x}$ is even and $h$ is even.

I claim that $h$ is infinitely differentiable, with

$$h^{(k)}(x) = \begin{cases}\frac{1}{2x^{k}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) & x \not= 0\\ \frac{f^{(k+2)}(0)}{2(k+1)} & x=0\end{cases}$$

In the base case, the identity holds for $k=0$. Now, assuming the identity holds for $k$, we have that for $x\not=0$

$$h^{(k+1)}(x) = \frac{-k}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) + \frac{x}{2x^{k+1}}\sum_{n=1}^k (-1)^{n+k} \frac{(k-1)!}{(n-1)!} x^{n-1} f^{(n+1)}(x)$$ $$+ \frac{x}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+2)}(x)$$ $$= \frac{1}{2x^{k+1}}\sum_{n=0}^{k+1} (-1)^{n+k+1} \frac{k!}{n!} x^n f^{(n+1)}(x)$$ Since the terms $n=1,2,\ldots k$ in the second sum cancel with the terms $n=0,1,\ldots,k-1$ in the third sum. And for $x=0$, we Taylor series expand each of the $f^{(m)}$ terms in $h^{(k)}$ to find that for $x \not=0$,

\begin{align*} h^{(k)}(x) &= \frac{1}{2x^{k+1}} \sum_{n=0}^{k} (-1)^{n+k} \frac{k!}{n!} x^n f^{(n+1)}(x)\\ &= \frac{(-1)^k}{2x^{k+1}} \sum_{n=0}^k (-1)^n \frac{k!}{n!} x^n \sum_{m=n}{^k+2} \frac{f^{(m+1)}(0)x^{m-n}}{(m-n)!} + o(x^{k+2})\\ &= \frac{(-1)^k}{2x^{k+1}}\left(\sum{m=0}^{k+2} f^{(m+1)}(0)x^m \frac{k!}{m!} \sum_{n=0}^m \frac{(-1)^n m!}{n!(m-n)!}\right) + \frac{f^{(k+2)}(0)}{2(k+1)} \\&\qquad+ \frac{xf^{(k+3)}(0)}{2}\left[\frac{1}{k+1} - \frac{1}{(k+1)(k+2)}\right] + o(x)\\ &= \frac{f^{(k+2)}(0)}{2(k+1)} + \frac{xf^{(k+3)}(0)}{2(k+2)} + o(x) \end{align*}

Thus, $$h^{(k+1)}(0) = \lim_{x\to0}\frac{h^{(k)}(x) - h^{(k)}(0)}{x} = \frac{f^{(k+3)}(0)}{2(k+2)}$$

This establishes the desired result.

• I don't think that your formula for $h'$ is correct. Also, it would probably be easier to begin the induction for $h^{(k)}$ by $k=0$, but your formula does not hold for this case. Feb 1, 2015 at 9:36
• @PhoemueX I've edited my answer.
– user88319
Feb 1, 2015 at 23:28
• looks like I have understood what you mean, thank you for your idea! Feb 2, 2015 at 6:12