# 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 $$C^1$$)