- #1

- 1,031

- 83

## Homework Statement

"Let ##f:ℝ^2\rightarrow ℝ## be a continuously differentiable function. Show that ##f## is not one-to-one."

## Homework Equations

A function ##f:ℝ^n\rightarrow ℝ^m## is

__continuously differentiable__if all the partial derivatives of all the components of ##f## exist and are continuous.

Hint: If, for example, ##D_1f(x,y)\neq 0## for all ##(x,y)\in A\subset ℝ^2##, consider ##g:A\rightarrow ℝ^2## defined by ##g(x,y)=(f(x,y),y)##.

Notation: ##D_1f(x,y)=\frac{d}{dx}f(x,y)##

## The Attempt at a Solution

I'm at a loss as to what to do here... I've tried finding ##Dg(x,y)## and then ##D(fg)(x,y)##, but I cannot seem to figure out a way to show that whenever ##(x,y_1)\neq (x,y_2)##, then ##f(x,y_1)= f(x,y_2)##. In order to do that, I plan on showing that ##D_2f(x,y)=0##. So thus far, I have:

##Dg(x,y)=\begin{pmatrix}

D_1f(x,y) & D_2f(x,y) \\

0 & 1 \\

\end{pmatrix}

\begin{pmatrix}

x \\

y \\

\end{pmatrix}##

So: ##D(fg)(x,y)=Df(g(x,y))Dg(x,y)=\begin{pmatrix}

D_1f(f(x,y),y) & D_2f(f(x,y),y) \\

\end{pmatrix}

\begin{pmatrix}

D_1f(x,y) & D_2f(x,y) \\

0 & 1 \\

\end{pmatrix}

\begin{pmatrix}

x \\

y \\

\end{pmatrix}=

xD_1f(x,y)D_1f(f(x,y),y)+y[D_1f(f(x,y),y)D_2f(x,y)+D_2f(f(x,y),y)]##

So I either must show that ##D_2f(x,y)=0##, or I'm approaching this problem incorrectly. Any insight on this will be very much appreciated. This is my most difficult class this semester, and it takes a while for me to pick up a new topic and learn it.

Last edited: