The following is an exercise from Calculus on Manifolds by Spivak.
Let $g$ be a continuous, real-valued function on the unit circle $\{x\in\mathbb{R}^2:\|x\|=1\}$, such that $g(-x)=-g(x)$.
Now define $f:\mathbb{R}^2\to\mathbb{R}$ by
$$f(x) =\begin{cases}
\|x\|\cdot g\Big(\frac{x}{\|x\|}\Big) & x\neq 0
\\ 0 & x=0
\end{cases}$$
Fix $x\in\mathbb{R}^2$, and further define a function $h:\mathbb{R}\to\mathbb{R}$ by $h(t)=f(tx)$.
Show that $h$ is differentiable.
My attempt:
Observe that for $t\neq 0$, we have
\begin{align*}
h(t)=f(tx) & = \|tx\|\cdot g\Bigg(\frac{tx}{\|tx\|}\Bigg)
\\ & = t\|x\|\cdot g\Bigg(\frac{tx}{t\|x\|}\Bigg)
\\ & = t\|x\|\cdot g\Bigg(\frac{x}{\|x\|}\Bigg)
\\ & = tf(x)
\end{align*}
In the case $t=0$, we have
$$h(0) = f(0) = 0$$
Thus, $h(t)=tf(x)$ for all $t\in\mathbb{R}$.
A function $f:\mathbb{R}\to\mathbb{R}$ is differentiable at a point $a\in\mathbb{R}$ if there exists a linear transformation $\lambda(h):\mathbb{R}\to\mathbb{R}$ such that
$$\lim_{h\to 0}\frac{f(a+h)-f(a)-\lambda(h)}{h}=0$$
Thus $h$ is differentiable if for all $t\in\mathbb{R}$, there exists a transformation $Dh(k):\mathbb{R}\to\mathbb{R}$ such that
$$\lim_{k\to 0}\frac{h(t+k)-h(t)-Dh(k)}{k}=0$$
We then have
\begin{align*}
\lim_{k\to 0}\frac{h(t+k)-h(t)-Dh(k)}{k} & =\lim_{k\to 0}\frac{(t+k)f(x)-tf(x)-Dh(k)}{k}
\\ & = \lim_{k\to 0}\frac{kf(x)-Dh(k)}{k}
\end{align*}
Thus the natural choice for $Dh(k)$ is $kf(x)$.
This completes the proof.
Questions:
- Does this proof work?
- How do I know that $Dh(k)=kf(x)$ is a linear transformation?
Best Answer
Certainly the idea of the proof is correct. What you want to end up showing is that for all $t \in \mathbb{R}$, and for all $x \in \mathbb{R^2}$, we have \begin{align} h(t) = t f(x). \tag{*} \end{align}
However, this requires some case work, which I'll outline for you, but I'll let you fill in the details.
Case $1$: $x = 0$.
(You supply the proof of why (*) is true)
Case $2$: $x \neq 0$. Here we have $3$ subcases:
Once you do this, then, you will have shown that for all $x \in \mathbb{R^2}$, $h: \Bbb{R} \to \Bbb{R}$ is a linear transformation. By the way, there are a few mistakes in your definition of differentiability. I'm not sure if these are typos or conceptual mistakes, but I'll highlight them anyway. First, you wrote
It should just say $\lambda : \Bbb{R} \to \Bbb{R}$, NOT $\lambda(h)$. $\lambda$ is the linear transformation, while $\lambda(h)$ is the real number you get when you evaluate $\lambda$ at a point $h$ in its domain. This kind of mistake appears again. You wrote:
The correct statement is: $h$ is differentiable if for every $t \in \mathbb{R}$, there is a linear transformation $Dh(t) : \mathbb{R} \to \mathbb{R}$, such that \begin{equation} \lim_{k \to 0} \dfrac{h(t+k) - h(t) - Dh(t)(k)}{k} = 0 \end{equation}
Notice that the derivative at $t$ is denoted $Dh(t)$. This itself is a linear transformation from $\Bbb{R}$ into $\Bbb{R}$. So you have to evaluate this on $k$. I suggest you read the definition of differentiability again, and pattern match what kind of object everything is.
Also, the correct answer for the derivative of $h$ is: for every $t \in \Bbb{R}$, $Dh(t)(\cdot) = h(\cdot)$, because $h$ is a linear transformation, as shown by equation (*) (I'd suggest you take a look at Theorem $2$-$3$ in the next section)
Added:
In this special case, as mentioned in the comments, the domain and target space of $h$ are $\mathbb{R}$. So you can use the old (single variable) definition of differentiability. You should also try to see why existence of $h'(t)$ (using single variable definition) and existence of $Dh(t)$ (using the definition above) are equivalent, and prove that $h'(t) = \left( Dh(t) \right)(1)$. (This is true in general for any function $h: \mathbb{R} \to \Bbb{R}$, not just the one defined in this question)