0
$\begingroup$

Suppose we have $f:I \to \mathbb{R^n}$, where $I \subset \mathbb{R}$ is in open interval, a continuously differentiable function.

If we define $g:I^2 \to \mathbb{R^n}$ by:

$g(x,y) = \begin{cases} \frac{f(x)-f(y)}{x-y} & x\neq y \\ Df(x)& x=y \end{cases} $

can you give me a hint as to how to show that $g$ is continuously differentiable on $I^2$\ $\{(x,x) | x \in I\}$?

I've showed that $g$ is continuous and I suspect that $Dg(x,y) = 0$, but haven't been able to show this using some forms of the mean value theorem, nor by looking at the strict definition of differentiability.

I also attempted to show that the partial derivatives of $g$ are continuous, with no success.

  • 0
    Choose $(a,b)\in I^2, a \ne b.$ Then there exists $h>0$ such that $b \notin (a-h,a+h).$ On this interval the function $x\to g(x,b)$ is the quotient of two continuously differentiable functions of $x.$ So now just march on ahead and find $\partial g /\partial x (a,b)$ using ordinary rules of calculus. – 2017-01-17
  • 0
    @zhw thanks for your comment; $g$ is a function whose image is in $\mathbb{R^n}$ - why would it partial derivative be defined? – 2017-01-17
  • 0
    @zhw.finding $D_x g_i$ for each $1 \leq i \leq n$ to be continuous seems to work to show $Dg$ is differentiable, but does it show that $Dg$ is continuously differentiable? – 2017-01-18

0 Answers 0