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.