# Suppose $f(t) = f(-t)$ and $f \in C^\infty$, show that there's a $h(t) \in C^\infty$ with $h(t^2) = f(t)$

So the hint of the exercise is that I should use Taylor's theorem (I'm not assuming $$f$$ is analytic).

Then it's clear that all odd derivatives of $$f$$ at $$0$$ is $$0$$ so in some sense $$h$$ can be defined by the Taylor coefficients $$h_k = f_{2k}$$.

More generally, can I show that if $$f \in C^{2k}$$ and is even, then there exists $$h \in C^k$$ with $$h(t^2) = f(t)$$?

• $\forall x>0\quad h(x):=f(\sqrt x).$ Commented Aug 20, 2023 at 21:39
• @AnneBauval Thank you for your response, I think it's clear that that's how you would define $h$, what's not clear to me is why $h \in C^k$ if $f \in C^{2k}$. Commented Aug 20, 2023 at 21:43
• At least on $(0,+\infty),$ since $\sqrt{}$ is smooth, $f\circ\sqrt{}$ has the same regularity as $f.$ Commented Aug 20, 2023 at 21:48
• You are right to be worried about the Taylor approach. If you take the (in)famous function $$f(x)=\begin{cases} e^{-1/x^2}, & x\ne 0 \\ 0, & x=0\end{cases},$$ then the Taylor series at $0$ is the zero function. Nevertheless, $h$ does exist. Commented Aug 21, 2023 at 2:21
• This was asked in 2011 on MO: mathoverflow.net/questions/72497/…
– KCd
Commented Aug 26, 2023 at 2:36

Here's an answer I have came up with using the idea suggested by copper hat in terms of looking at decay of the remainder term. I am not sure if $$h(t) =: f(\sqrt t)$$ is actually $$C^k$$ if $$f \in C^{2k}$$ However, I claim that

Suppose $$f$$ is even and $$f \in C^{3k+1}(\mathbb{R})$$, then $$h(t)=: f(\sqrt t) \in C^k([0,\infty))$$

Proof: Apply Taylor's theorem to $$f$$ at $$0$$ to the $$2k$$ order (assuming $$k \geq 1$$.

Then $$f(t) = \sum_0^{k} a_i t^{2i} + t^{2k+1}\int_0^1 f^{(2k+1)}(ts)(1-s)^{2k+1} ds$$

Then for $$t \geq 0$$, $$h(t) = f(\sqrt t) = \sum_0^k a_i t^i + t^{\frac{2k+1}{2}}\int_0^1 f^{(2k+1)}(\sqrt{t}s)(1-s)^{2k+1} ds$$

Now for any $$l \leq k$$, if we take $$l$$th derivative then the remainder term will involve a positive power of $$t$$ times an integral expression with bounded integrand since we can still take up to $$k$$ derivative of $$f^{(2k+1)}$$. This can be seen using chain rule and product rule and differentiation under the integral. (Edit: To elaborate, when differentiating the remainder term, any derivative of the intgrand will decreases the total power of t of the remainder by $$\frac{1}{2}$$ by chain rule, while differentiating $$t^{\frac{2k+1}{2}}$$ will decrease that power by $$1$$, since I'm not differentiating more than $$k$$ times the total power of $$t$$ from differentiating the remainder term will be positive, hence tends to 0 as $$t$$ tends to 0)

So $$\lim_{t \rightarrow 0^+} h^{(l)}(t)$$ can be read off from the $$l$$th term in the polynomial part (the first sum)

QED

It is known that for every sequence there exists a smooth function whose Taylor series coefficients are equal to that sequence, consider the sequence with the first $$k$$ terms are $$\lim_{t \rightarrow 0^+} \frac{1}{l!}h^{(l)}(t)$$, then define $$h(t)$$ to be equal to the said function above for $$t \leq 0$$, then this $$h$$ would have the desired property $$h(t^2) = f(t)$$ and it is $$C^k$$.

Remark: I'm not sure if this is the sharpest result, but it is enough to show that if $$f \in C^\infty$$, then there exists such a $$h \in C^\infty$$.

• When you differentiate the integral remainder term, you cannot ignore the $\sqrt{t}$ term, which is what I did implicitly in my deleted answer. See @AnneBauval's comment. Commented Aug 21, 2023 at 6:26
• But I haven't ignored the $\sqrt(t)$ term, when you differentiate the integrand with respect to $t$, you get $\frac{1}{2\sqrt{t}}f^{m+1}(\sqrt{t}s)$, which decreases the power of $t$ for the whole remainder term by $\frac{1}{2}$, but since I am not taking more than k derivatives I have "enough" power of t that will go to 0. The need to differentiate the integrand is also why I'm requiring f to be 3k+1 differentiable. Commented Aug 21, 2023 at 6:35
• Correction, there should also be a factor of s from chain rule. Commented Aug 21, 2023 at 6:44

Let us prove that if $$k\ge1$$ and $$f:\Bbb R\to\Bbb R$$ is even and $$2k$$ times differentiable (not necessarily $$C^{2k}$$), then $$f(x)=h(x^2)$$ for some $$C^k$$ function $$h$$ on $$[0,+\infty)$$ (which can then be extended to a $$C^k$$ function on $$\Bbb R,$$ thanks to Borel's lemma). The only problem is the regularity of $$h$$ at $$0.$$

For a fixed $$k,$$ consider the universal constants $$a_0,\dots,a_k$$ (we won't need to calculate them) such that for every $$k$$ times differentiable function $$F:[0,+\infty)\to\Bbb R,$$ the function $$H:[0,+\infty)\to\Bbb R$$ defined by $$H(x^2)=F(x)$$ satisfies: $$\forall x>0\quad H^{(k)}(x^2)=\frac1{x^{2k}}\sum_{i=0}^ka_ix^iF^{(i)}(x).$$ If $$F$$ is moreover $$2k$$ times differentiable, Taylor gives: \begin{align}H^{(k)}(x^2)&=\frac1{x^{2k}}\left(\sum_{i=0}^ka_ix^i\sum_{j=0}^{2k-i}\frac{F^{(i+j)}(0)}{j!}x^j+o(x^{2k})\right)\\ &=\frac1{x^{2k}}\left(\sum_{n=0}^{2k}c_nF^{(n)}(0)x^n+o(x^{2k})\right) \end{align} where the (again universal!) constants $$c_0,\dots,c_{2k}$$ are given by: $$c_n=\sum_{i\le\min(k,n)}\frac{a_i}{(n-i)!}.$$ Now, since $$f$$ was assumed to be even, its derivatives of odd order vanish at $$0,$$ so that for our function $$h(t):=f(\sqrt t),$$ the previous identity simplifies to $$\forall x>0\quad h^{(k)}(x^2)=\frac1{x^{2k}}\left(\sum_{m=0}^kc_{2m}f^{(2m)}(0)x^{2m}+o(x^{2k})\right)$$ Assuming (by induction) that we already know that $$h$$ is $$C^{k-1},$$ we are done if we prove that $$\forall m because this will entail the existence of $$\lim_{0^+}h^{(k)}$$ (equal to $$c_{2k}f^{(2k)}(0)$$), hence the continuous differentiability at $$0^+$$ of $$h^{(k-1)}.$$

Here comes the magic of the universality: to prove that $$c_{2m}=0,$$ simply apply the formula above to some $$C^\infty$$ function $$h$$ for which the $$2m$$-th differential at $$0$$ of $$x\mapsto h(x^2)$$ is non-zero (I leave it to you to produce such a function $$h$$), and use that $$\lim_{x\to0}h^{(k)}(x^2)$$ exists.

• I'm having a hard time understanding this proof, what are universal constants (forgive my ignorance). And also I'm not seeing why you can put $H(x^2)$ in that form to begin with. Commented Aug 29, 2023 at 7:42
• @Ecotistician Universal means not depending on the functions you consider. For instance $H(x^2)=F(x)\implies 2xH'(x^2)=F'(x)$ hence if $k=1$ then $a_0=0$ and $a_1=\frac12,$ whatever function $F$ you chose. As for $H(x^2)=F(x):$ once you chose $F,$ simply define $H$ by $H(t):=F(\sqrt t)$ (for all $t\ge0$). Commented Aug 29, 2023 at 7:49
• I see what you are doing now, this is indeed a marvelous proof which gives the sharpest result. Thank you Anne! Edit: Although, are you sure we didn't need to assume that f is actually in $C^{2k}$ in order to apply taylor's theorem, and to use $lim_{0^+} h^{k}(x^2)$ exists? Commented Aug 29, 2023 at 12:21
• @Ecotistician Thank you for the praises! Yes I am sure: Taylor's theorem $f(x)=f(0)+\dots+\frac{f^{(n)}(0)}{n!}x^n+o(x^n)$ only needs $f$ to be $n$ times differentiable at $0,$ $f^{(n)}(x)$ does not even need to exist for $x\ne0.$ And the existence of $\lim_{0^+}h^{(k)}$ is sufficient for $h$ to be $C^k$ at $0^+,$ if $h$ is already known to be $C^{k-1}.$ Commented Aug 29, 2023 at 12:45
• May I ask how did you come up with this approach? What were your train of thoughts that lead you to try this? I understand the proof line by line but I'm not seeing what the underlying idea is. Commented Aug 29, 2023 at 23:06

It is clear that for $$t \ge 0$$ that we must have $$h(t) = f( \sqrt{t})$$. Since $$f$$ is smooth and $$t \mapsto \sqrt{t}$$ is smooth for $$t>0$$, we see that $$h$$ is smooth on $$t >0$$.

Since $$f$$ is even, for any $$n$$, Taylor's theorem gives $$f(t) = \sum_{k=0}^n {f^{(2k)}(0) \over (2k)!} t^{2k} +R_{2n+1}(t)$$, where $$R_k(t) = {t^{k+1} \over k!} \int_0^1 f^{(k+1)}(st) (1-s)^k ds$$. Hence, for $$t>0$$ we have $$h(t) = \sum_{k=0}^n {f^{(2k)}(0) \over (2k)!} t^{k} +R_{2n+1}(\sqrt{t})$$. (Note that the subscript on $$R$$ is $$2n+1$$.)

(Thanks to @AnneBauval's meticulous attention for catching an overreaching conclusion at this point in an earlier version of the proof.)

Let $$r(t) = R_{2n+1}(\sqrt{t})$$, then $$r(t) = {t^{n+1} \over (2n+1)!} \int_0^1 f^{(2n+2)}(s\sqrt{t}) (1-s)^{2n+1} ds$$.

If we show that the one sided derivatives $$r^{(k)}(0)$$ exist and are zero for $$k=0,...,n-1$$ then it follows that the same is true for $$h^{(k)}(0)$$. Since $$n$$ is arbitrary, if follows that $$h^{(k)}(0)$$ exists for all $$k$$.

Let $${\cal T}_p$$ be the span of the functions of the form $$t \mapsto t^l \int_0^1 f^{(k+1)}(s\sqrt{t}) (1-s)^{k} s^m ds$$, for $$k \ge 0, l \ge p, m \ge 0$$ (and $$t \ge 0$$, of course). The elements of $${\cal T}_p$$ are continuous for $$p \ge 0$$.

If $$p>0$$ and $$\phi \in {\cal T}_p$$, note that $$\phi(0) = 0$$. If $$p>1$$ and $$\phi \in {\cal T}_p$$, note that $$\phi$$ is differentiable (a one sided derivative from the right at $$t=0$$), $$\phi'(0) = 0$$ and $$\phi' \in {\cal T}_{p-1}$$.

Since $$r \in {\cal T}_{n+1}$$, it follows that $$r(0)=r^{(1)}(0) = \cdots = r^{(n-1)}(0) = 0$$ and so $$h$$ has one sided derivatives at $$t=0$$ and these are given by $$h^{(k)}(0) = k!{f^{(2k)}(0) \over (2k)!}$$.

All that remains is to show that $$h$$ can be extended to a smooth function for $$t <0$$. The tool for this is Borel's Lemma applied to the sequence $$h^{(k)}(0)$$. This gives the existence of a smooth function $$g$$ such that $$g^{(k)}(0)= h^{(k)}(0)$$.

To finish, let $$h(t) = \begin{cases} f(\sqrt{t}, & t>0,\\ g(t),& \text{otherwise}\end{cases}$$

• It does not "follow from this that $h$ has (at least) one sided derivatives at $t=0$". Look at the counterexample at the end of this paragraph. Commented Aug 21, 2023 at 4:46
• @AnneBauval Thanks for catching that. Commented Aug 21, 2023 at 5:26
• Thank you copper hat, does this proof assume f is at least 2k+2 (or 2k+1) differentaible then ? Commented Aug 29, 2023 at 7:42
• I am not sure what you are asking. $f$ is assumed smooth. Commented Aug 29, 2023 at 7:55