Given $f:\mathbb R^2 \to \mathbb R$, and assuming that the directional derivative $D_vf(x)$ is uniformly continuous in $x$ for $|v|=1$, show that $f \in C^1$.
This is an old exam question dating back to January 1990 that I would like to work through.
There is a hint given for this problem:
Use the Fundamental Theorem of Calculus to show that
$$f(x+h) - f(x) - D_hf(x) = \int_0^1 D_hf(x+sh)-D_hf(x) \, ds$$
Some initial thoughts:
I need to show that f is both (totally) differentiable and the derivative $f'$ is continuous. So, I need to show that $f$ is differentiable for arbitrary $x$ in its domain, i.e., show that
$$\lim_{h\to 0} \frac {|f(x+h) - f(x) - Ah|}{|h|} =0,$$
where A is a linear operator mapping $\mathbb R^2 \to \mathbb R$.
Is this "$A$" always the derivative matrix? The definitions seem very technical and abstract and doesn't say it explicitly -- the definitions just say "there exists a linear operator A such that ... the above limit is zero."
If $A$ is the derivative matrix, then for this problem, it would be a 1x2 matrix, i.e., the gradient of $f$.
So, the numerator in the limit looks pretty close to the left-hand side of the hint.
Also, $D_vf(x)$ is uniformly continuous in $x$, so
for $\epsilon$ >0, there exists $\delta>0$, such that $|x-y|< \delta$
implies that $|D_vf(x) - D_vf(y)| < \epsilon$.
What does uniformly continuous in $x$ for $|v| = 1$ mean? Does this mean that $D_vf(x)$ is uniformly continuous on the unit circle only and not in the interior?
Any suggestions on how to get started on this problem with the above information is welcome -- I'd also welcome any solutions, too.
Thanks,
Yes, $A$ is the derivative matrix. Apparently your book is a little unclear on this, but because you can't divide by the vector $h$, the more familiar form of the definition of derivative doesn't work. It had to be recast in a form where the only division going on is by the scalar $|h|$.
And yes, you are only know that $D_vf(x)$ is uniformly continuous on the boundary of the unit circle, not the interior.