# How to show that this function $f$ is continuously differentiable,

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.
• To clarify a bit: there is nothing said about the interior. $D_vf(x)$ could be unitformly continuous on the interior, or not. – Paul Sinclair Sep 25 '15 at 2:19
• Hi @PaulSinclair, I am working through it now - seems a bit scary and technical, but so far it's been manageable. But, to use the FTC, can I assume that $D_vf(x)$ is continuous from 0 to 1? I'm a little confused about what assumptions I can make about the directional derivative. If it is uniformly continuous on |v| =1, must it be at least continuous from 0 to 1, so that I can integrate it? It doesn't seem obvious... – User001 Sep 25 '15 at 3:27
• The directional derivative is linear in $v$. In particular, $D_{av}f(x) = aD_vF(x)$ (Which BTW belies my earlier comments, but then my point then was that you weren't given this information, not that it necessarily was false.) – Paul Sinclair Sep 25 '15 at 3:41
• Always $D_vf(x) = Df|_x v$. That is, the directional derivative in direction $v$ at the point $x$ is the total derivative matrix at $x$ applied to the vector $v$ – Paul Sinclair Sep 25 '15 at 16:22
• It is not a gradient as such. In general, the derivative of a function $f$ from $\Bbb R^m \to \Bbb R^n$ at a point $x$ is a linear operator - an $m \times n$ matrix. If $\phi$ is a curve in $\Bbb R^m$ that at $x$ is tangent to a vector $v$, then $f\circ\phi$ is a curve in $\Bbb R^n$ which at $f(x)$ is tangent to the vector $Df|_xv$. In the special case $n = 1$, the derivative is a linear functional from $\Bbb R^m \to \Bbb R$, but by the inner product on $\Bbb R^m$, this functional can be converted to a vector. That vector is the gradient. – Paul Sinclair Sep 26 '15 at 4:29