Let $F: \mathbb R^m \to \mathbb R^n$ be smooth, $F(0) = 0$. Let $H(t,x) = \frac{F(tx)}{t}$ if $t \neq 0$ and $DF|_0(x)$ otherwise. Show $H$ is smooth

Question

This question is taken from my differential topology lecture notes.

Let $F : \mathbb R^m \to \mathbb R^n$ be a smooth map such that $F(0) = 0$ and define $H: \mathbb R^{m+1} \to \mathbb R^n$ as $$H(t,x) := \begin{cases} t^{-1}F(tx) & \text{ if } t \neq 0 \\ DF|_0 (x) & \text{ if } t = 0 \end{cases}.$$

Show that $H$ is a smooth function.

Of course, when $t \neq 0$, $H$ is easily seen to be smooth. The question is about proving that $H$ s a smooth function when $t = 0$. I think I can show that $\frac{\partial H}{\partial t}$ is a smooth function for $t = 0$ and $x = \tilde{x}$ arbitrary through the use of L'Hopital's rule and using the definition of a partial derivative, but I am not sure how I can generalize to show that $H$ is a smooth function. Someone suggested that I try to come up with a function $g$ that takes as input a real number and such that $F'(tx) = g'(1) - g'(0) = \int_0^1 g''(x) dx$ which does not make sense to me. I also am thinking about using Taylor's series, but it seems a little complicated. Is there an easy way to see that $H$ is smooth at $t = 0$ and $x$ arbitrary? A similar but not the same question has been asked here whose answer did not make sense to me. Is there another way to solve this?

Answer 1

Consider the function $G(t,x):=\int_0^1DF_{tsx}(x)\,ds$. This is a $C^{\infty}$ function because $F$ is smooth (so $(t,s,x)\mapsto DF_{tsx}(x)$ is smooth) and the integration preserves smoothness (this is a very standard lemma, which I leave to you to prove).

The claim is that this function $G$ is actually equal to your $H$.

• Notice that $G(0,x)=\int_0^1DF_0(x)\,ds=DF_0(x)=H(0,x)$.
• Next, notice that by the fundamental theorem of calculus, \begin{align} F(tx)&=F(0)+\int_0^1\frac{d}{ds}\left(F(tsx)\right)\,ds\tag{FTC}\\ &=0+\int_0^1DF_{tsx}(tx)\,ds\tag{chain rule}\\ &=t\int_0^1DF_{tsx}(x)\,ds\tag{$DF_{tsx}(\cdot)$ is a linear map}\\ &=t\cdot G(t,x) \end{align} so for $t\neq 0$, we have $G(t,x)=\frac{F(tx)}{t}=H(t,x)$.

This proves $H=G$, and we already know $G$ is $C^{\infty}$, so $H$ is $C^{\infty}$.

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This is of course very closely related to the comment, because $G(t,x)=\int_0^1DF_{tsx}(x)\,ds$ can also be written (after writing the total derivative in terms of partials) as $\sum_{i=1}^nx^i\int_0^1\frac{\partial F}{\partial x^i}(tsx)\,ds$.