1. Oct 21, 2013

Felafel

1. The problem statement, all variables and given/known data

f(x,y) is a two variables function for which the hypothesis of Schwarz's theorem hold in a point (a,b).
is f continuous in (a,b)?

3. The attempt at a solution

I think it is, because being the two mixed partials continuous in (a,b) the function is twice differentiable, and therefore globally Lipschitz, which implies continuity.
Am I right?

2. Oct 21, 2013

brmath

Are you referring to the Schwarz' theorem that says if $f_{xy}$ and $f_{yx}$ are both continuous at point (a,b) there is a neighborhood U of (a,b,) where$f, f_x, f_y,f_{xy}, f_{yx}$ all exist and $f_{xy} = f_{yx}$ at (a,b)?

We cannot infer from this that f is continuous at (a,b). That seems unintuitive, because if you said the same for a function of a single variable, of course it would be continuous.

The problem is that existence or continuity of the partial derivatives is nowhere near as strong as existence of a derivative of a function of 1 variable. The partials only tell you that there is continuity along two different lines coming into (a,b). The function could fail to be continuous at (a,b) because it has a discontinuity there coming in along some other path.

Remember that continuity at (a,b) means that for any $\epsilon$ > 0 there exists some neighborhood U of (a,b) where (x,y) $\in$ U $\Rightarrow$ |f(x,y)-f(a,b)| < $\epsilon$.

Okay, an example. Let (a,b) = (0,0). Let f(x,y) = x when x = 0 and y when y = 0, but f(x,y) = 2x +2y + 1 if x and y are not zero. Then $f_x$(0) = $f_y$(0) = 1 and $f_{xy}(0)$ = $f_{yx}(0) = 0.$ But as per the definition above, f is not continuous at 0.

I believe part of the point of this problem is precisely to show that partial derivatives are not analogous to the derivative in one dimension. There is in fact an analogy in multiple dimensions -- you can look up what "differentiability" means for functions of multiple variables.

3. Oct 22, 2013

Felafel

super! thank you very much :D