Is this derivative Lipschitz?

1. Jul 21, 2009

WastedGunner

Here is a tough one:

Say we have a multivariable function f:R^n -> R
and for any x, and direction u, the function g:R->R defined as g(t)=f(x+tu) has that g'(t) is Lipschitz with the same Lipschitz constant (say M). For special cases, taking u to be any basis element we see that every partial restricted to the direction of its axis is Lipschitz

Can I then prove that the derivative Df(x) (or equivalently any partial derivative) is Lipschitz as well.

the 1-dimensional case is trivial:

|(Df(y)-Df(x))\cdot e_i| = |g'(y)-g'(x)|<M|y-x| Lipschitz. (g is defined above with x=0 and u=e_1)

It is the n>1 cases that poses a problem, because I find that if y-x is not parallel with e_i we run into problems.

I have been working on this problem for a while and find I am working in circles now.

2. Jul 21, 2009

jostpuur

You know sometimes people think that if partial derivatives exist, then the function would be differentiable, but that's not always true, because it can happen that the expression

$$\frac{f(x+tu) - f(x)}{t}$$

has a limit when $t\to 0$ for all fixed $u\in\mathbb{R}^n$, but for some suitable differentiable function $u(t)=a_1t + \frac{1}{2}a_2t^2 + \cdots$, a limit of an expression

$$\frac{f(x+u(t)) - f(x)}{t}$$

(if existing) does not agree with the partial derivative in direction $a_1$.

Perhaps a similar problem could ruin your hypothesis? There might exist a differentiable function, for which the derivatives in any fixed direction are always Lipschitz, but for which derivatives along some curves are not?

3. Jul 22, 2009

WastedGunner

Also, on the note of partial derivatives, it is true however that if all the partials are continuous that the function is differentiable (in fact it is $$C^1$$)