Question about the implicit function theorem

Question

Define $f(u,v) = w$ for $u,v,w \in \mathbb{R}$, $f \in C^1$ . I'm told that $f(0,0) = 0 $ and that $af_u(0,0) + bf_v(0,0) \neq 0$.

I'm asked to prove: the equation $f(x-az,y-bz) = 0 $ defines $z$ as a function of $x,y$ for some space near $(0,0,0)$ in $\mathbb{R}^3$.

I know that the solution to this problem is through the implicit function theorem: but in order to use the theorem I need to check that the condition $f_z(0,0) \neq 0$ and this is where I can't seem to understand the problem. How did we get to $\mathbb{R}^3$? Do I need to derive a composite function? I don't know how to approach this problem.

Any help would be appreciated!

Answer 1

My interpretation of the problem is the following.

Let $f\colon \mathbb R^2\to \mathbb R, (u,v)\mapsto f(u,v)$ be a class $C^1$ function such that $f(0,0)=0$ and assume that $a,b$ are real numbers such that $af_u(0,0)+bf_v(0,0)\neq 0.$

Prove that the equation $\varphi(x,y,z)=0$, where $\varphi\colon \mathbb R^3\to \mathbb R, (x,y,z)\mapsto f(x-az, y-bz)$, defines $z$ as a function of $(x,y)$ around $(0,0,0)$.

In the notation of the implicit function theorem's link on wikipedia one has $\varphi$ instead of wikipedia's $f$, plus $n=2, m=1, (a_1, a_2, b_1)=(0,0,0)$ and $c=0$. Obviously $\varphi$ is continuously differentiable so in order to apply the theorem you need to check that

• $\varphi(0,0,0)=0$;
• $\varphi_z(0,0,0)\neq 0$.

To check that $\varphi_z(0,0,0)\neq 0$ holds use the chain rule which tells you that for all $(x,y,z)\in \mathbb R^3$, $$\varphi_z(x,y,z)=f_u(x-az,y-bz)u_z(x,y,z)+f_v(x-az, y-bz)v_z(x,y,z),$$ where $u\colon \mathbb R^3\to \mathbb R, (x,y,z)\mapsto x-az$ and $v\colon \mathbb R^3\to \mathbb R, (x,y,z)\mapsto y-bz$.

The symbol $\varphi_z$ should be interpreted at the partial derivative of $\varphi$ with respect to its third variable. Similarly $f_u$ and $f_v$ should be interpreted as the partial derivatives of $f$ with respect to its first and second variables, respectively.