Show that a $f(r\cos\theta, r\sin\theta)=rh(\theta)$ is continuous on $\mathbb{R}^2$. "Vector Calculus ..." - Hubbard & Hubbard: Exercise 1.20

There's this exercise in Hubbard's book:

Let $h:\Bbb R \to \Bbb R$ be a $C^1$ function, periodic of period $2\pi$, and define the function $f:\Bbb R^2 \to \Bbb R$ by $$f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=rh(\theta)$$

a. Show that $f$ is a continuous real-valued function on $\Bbb R^2$.

b. Show that $f$ is differentiable on $\Bbb R^2 - \{\mathbf 0\}$.

c. Show that all directional derivatives of $f$ exist at $\mathbf 0$ if and only if

$$h(\theta) = -h(\theta + \pi) \ \text{ for all } \theta$$

d. Show that $f$ is differentiable at $\mathbf 0$ if an only if $h(\theta)=a \cos \theta + b \sin \theta$ for some number $a$ and $b$.

I can't find how to prove $f$ is continuous, I tried to prove $$\lim_{\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix} \to \begin{pmatrix}s\cos\phi\\s\sin\phi \end{pmatrix}} f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=s\ h(\phi)$$ for all $s$ and $\phi$. But I can't do much else.

• I might try try $f(x,y) = \sqrt {x^2 + y^2} h(arctan \frac yx)$ Oct 5, 2016 at 20:08

First consider the limit of $f$ as you approach some point $\begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}$ for which $s>0$. \begin{align*} \lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}}f\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix}&= \lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} rh(\theta)\\ &=\left(\lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} r\right)\left( \lim_{\begin{pmatrix}r \cos \theta\\r \sin \theta\end{pmatrix} \to \begin{pmatrix}s \cos \phi\\s \sin \phi\end{pmatrix}} h(\theta)\right)\\ &= s\cdot h(\phi) \text{ (by continuity)}\\ &=f\begin{pmatrix}s\cos\phi \\ s \sin \phi\end{pmatrix}. \end{align*} Now if you are approaching the origin (which corresponds to $s=0$), $\theta$ need not approach a limit. To show this function is continuous at $(0,0)$, you need to show $$\lim_{r \to 0}f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=f\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ regardless of $\theta$. For this you should use the squeeze theorem. Note that since $h$ is continuous and periodic with period $2\pi$, $h$ is bounded on $[0,2\pi]$ and thus on all of $\mathbb{R}$.
• Thank you. I think I'd got confused by the way $f$ is defined. Oct 5, 2016 at 20:17
• (a) Observe \begin{align} |f(r\cos\theta, r\sin\theta) - f(s\cos\phi, s\sin\phi)| = |rh(\theta)-sh(\phi)| = |h'(\xi)(r-s)|\leq M|r-s| \end{align} since $h$ is periodic and $C^1$.
• (c) By the definition, you need to show \begin{align} \lim_{t\rightarrow 0} \frac{f(tm, tk)-f(0, 0)-t[\partial_xf(0,0)m+\partial_yf(0, 0)k]}{t} =0 \end{align} for all $(m, k)= (a\cos\phi, a\sin\phi)\neq(0, 0)$. Using the given relation, we have that \begin{align} &\frac{f(tm, tk)-f(0, 0) - t[\partial_xf(0,0)h+\partial_yf(0, 0)k]}{t} =\ \frac{at h(\phi)-t[\partial_xf(0,0)m+\partial_yf(0, 0)k]}{t}\\ &=\ ah(\phi) -a[\partial_xf(0,0)\cos\phi+\partial_yf(0, 0)\sin\phi] =0. \end{align} Then you need to show the above holds for all $(a\cos\theta, a\sin\theta)$ if and only if $h(\theta) = -h(\theta+\pi)$. Of course, you need to first check the partial derivatives at the origin exists and determine their values.