Is the Derivative of a Multivariable Function Lipschitz?

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The discussion centers on the Lipschitz continuity of the derivative of a multivariable function, specifically when the function f: R^n -> R has Lipschitz continuous directional derivatives g'(t) with a constant M. The user explores whether this property implies that the overall derivative Df(x) is also Lipschitz. While the one-dimensional case is straightforward, the user encounters challenges in higher dimensions (n > 1) due to potential non-parallelism of vectors. The existence of Lipschitz continuous partial derivatives does not guarantee differentiability, and the user acknowledges the complexity of constructing a counterexample.

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WastedGunner
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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.
 
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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

<br /> \frac{f(x+tu) - f(x)}{t}<br />

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

<br /> \frac{f(x+u(t)) - f(x)}{t}<br />

(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?
 
Thanks for your reply.

I know that the existence of partial derivatives does not guarantee a differentiable function.

I have been unable to construct a counter example, the condition that all those function have a uniform Lipschitz constant is a very strong condition.

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)

I guess I'll just think about this problem some more.

PS i read how to imbed latex into posts.
 

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