What does it really mean to say a function $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is continuously differentiable?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuously differentiable if $f$ is differentiable and $f':\mathbb{R}\rightarrow \mathbb{R}$ is continuous..
But what is $f'$ in case of $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$..
Is it the jacobian $\begin{bmatrix}\frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}& \cdots &\frac{\partial f_1}{\partial x_n} \\
\frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}& \cdots &\frac{\partial f_2}{\partial x_n} \\
\vdots \\
\frac{\partial f_n}{\partial x_1}& \frac{\partial f_n}{\partial x_2}& \cdots &\frac{\partial f_n}{\partial x_n} \end{bmatrix}$
I could not understand how does this matrix act on $\mathbb{R}^n$..
As an example, for $f(x,y)=(xy,x+y)$ we see that jacobian is
$\begin{bmatrix}y&x\\1&1\end{bmatrix}$.. I do not understand how do i define
$Df: \mathbb{R}^2\rightarrow \mathbb{R}^2$..
In case of derivative at a point we have $Df((a,b))=\begin{bmatrix}b&a\\1&1\end{bmatrix}$ and we define
$Df(a,b)(x,y)=\begin{bmatrix}b&a\\1&1\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}bx+ay\\x+y\end{bmatrix}$
I mean there are already variables in $Df$ and so i am getting confused how to act on $\mathbb{R}^n$.. But in case of $Df(a,b)$ there are no variables.. So, it seems to be natural.
Best Answer
Given a function $f:\>\Omega\to{\mathbb R}^m$ with domain $\Omega\subset{\mathbb R}^n$ its derivative is a map $$df:\quad\Omega\to{\cal L}({\mathbb R}^n,{\mathbb R}^m),\qquad x\mapsto df(x)\ .$$ The "coordinates" of $df(x)$ are the entries of the Jacobian $$J(x):=\left[\matrix{f_{1.1}&\cdots& f_{1.n}\cr\vdots\cr f_{m.1}&\cdots& f_{m.n}\cr}\right]_x\ ,\qquad f_{i.k}(x):={\partial f_i\over\partial x_k}(x)\ .$$ The space ${\cal L}({\mathbb R}^n,{\mathbb R}^m)$ has a natural metric induced by the norm $\|A\|:=\sup_{|x|=1}|Ax|$. It so happens that the function $df$ is continuous on $\Omega$ iff all entries $f_{i.k}$ of the Jacobian are continuous on $\Omega$. This fact belongs to elementary limit geometry in ${\mathbb R}^d$ and has nothing to do with differentiation per se.