Confusion with partial derivatives on open sets

The most of the times when we talk about the partial derivatives of a function $f:D\subset{R^2}\rightarrow{R}$ ,we ask for the domain $D$ to be an open set of $R^{2}$. In my opinion this is very restrictive so I think, in order to determine let's say the $\partial_1f(x_0,y_0)$it would suffice for the set $D$ to include an horizontal line segment containing the point $(x_0,y_0)$. For the following example, it would be: $$\lim_{t\to x_0}\frac{f(t,y_0)-f(x_0,y_0)}{t-x_0}=\partial_1f(x_0,y_0)$$ which actually means:$$\lim_{t\to x_0^+}\frac{f(t,y_0)-f(x_0,y_0)}{t-x_0}=\partial_1f(x_0,y_0)$$

The same thing, we may deduce for $\partial_2f(x_0,y_0)$ and so on. My question here is if I miss something or have something not well understood...Thanks. • If you are defining the function $f(x,y)$ over the triangle region in your picture, for which $(x_0,y_0)$ is the lower vertex, then $y_0$ is the smallest $y$-coordinate in the triangle and there is only one point in the triangle that has that $y$-coordinate, namely, $(x_0,y_0)$. So I do not understand your limit with $f(t,y_0)$ and $t\rightarrow x_0^+$ since such points $(t,y_0)$ are not in the domain of $f$. Jul 6 '18 at 19:58
• @Michael, you are right, the triangle was not a good idea. But how about a rectagular? Jul 6 '18 at 20:28
• Your one-sided limit definition would at least make sense for the rectangle, provided the rectangle is oriented the same as the $x-y$ axis and assuming you re-define the one-sided limit depending on where you are on the boundary (the right boundary would need a one-sided limit that approaches from the left, rather than the right). Jul 6 '18 at 20:37
• @Michael, and something more...what if the function $f$ is defined on the not open set $Q^2$? Then I think we can also talk about the $lim$:$$\lim_{t\to x_0}\frac{f(t,y_0)-f(x_0,y_0)}{t-x_0}=\partial_1f(x_0,y_0)$$ Jul 7 '18 at 6:28
• You can make any definition you want as long as it makes sense, is useful, and you are clear about it (particularly when the definition is not standard). One possible way to get a partial derivative for your triangle example, even at the bottom corner point, is if you can find a (unique?) vector $v$ such that $$\lim_{w\rightarrow z} \frac{||f(w)-f(z)-v\cdot (w-z)||}{||w-z||}=0$$ where the limit is taken over $w$ in the domain of $f$, then take dot products of $v$ with a unit vector in a direction of interest. Jul 7 '18 at 15:53

Define $$\mathcal{X} = \{(x,y) \in \mathbb{R}^2 : x=2y\}$$ The set $\mathcal{X}$ is not open. Define (convex) functions $f:\mathcal{X}\rightarrow\mathbb{R}$ and $g:\mathcal{X}\rightarrow\mathbb{R}$ by $$f(x,y) = x+y \quad, \quad g(x,y) = 3y$$ We might be tempted to say that the “gradients” of $f$ and $g$ are $$\nabla f(x,y) = [\partial f/\partial x; \partial f/\partial y] = [1; 1] \quad, \quad \nabla g(x,y) = [\partial g/\partial x; \partial g/\partial y] = [0; 3]$$ However, notice that, in fact, $f$ and $g$ are exactly the same function! So the notion of “gradient” is not necessarily clear (or unique) when the function is not defined over an open set.
On the other hand, both $[1;1]$ and $[0;3]$ are subgradients of the function (at every point of its domain $\mathcal{X}$).
• nice anti example, but as in my triangle the function $f(t,y_0)$ is not defined here in any interval containing $x_0$ Jul 6 '18 at 20:40