Originally Posted by

**arbolis** Let $\displaystyle A=\{ (x,y) \in \mathbb{R}^2 :y \leq x-1 \}$ and let $\displaystyle f(x,y)$ be defined in all $\displaystyle \mathbb{R}^2$ by $\displaystyle f(x,y)= \begin{cases} y^2-(x-1)^2 \hspace{2cm} \text{if (x,y)} \in A \\ 0 \hspace{3 cm} \text{if (x,y) is not in A} \end{cases}$.

1)Analyze the continuity of $\displaystyle f$ in all $\displaystyle \mathbb{R}^2$.

I'll put the other questions if I need help, but I'd like a confirmation of my result for part 1).

My attempt :

First I tried to visualize the function and it was quite hard. I couldn't even do that. I realize that it takes the element $\displaystyle y=x-1$ (a line) from the domain and send it to the image as the line $\displaystyle y=0$ (or curve?). I also realize that it sends all the elements under the curve $\displaystyle y=x-1$ into another area, but I didn't visualize it. Anyway, back to the question, I think the limit exists by intuition.

Here's my work to support this : $\displaystyle f$ is continuous in all $\displaystyle \mathbb{R}^2$ except maybe in $\displaystyle y=x-1$.

Right. But you could have been a bit more explicit here. For example you could have argued like this:

A is the half-plane below (and including) the line $\displaystyle y=x-1$. f happens to be continuous at any *inner* point of either A or $\displaystyle \mathbb{R}^2\backslash A$: because f happens to be identical to a continuous function in an entire neighborhood of such a point.

Thus, the only problematic points are those on the line $\displaystyle y=x-1$.

But note: by arguing like this you already assume that f restricted to A happens to be continuous, so that there is no reason to even try to prove it only for points on the line $\displaystyle y=x-1$ the way you do below:

I'll try to prove it is :

I want to prove that $\displaystyle \lim _{y \to x-1} f(x,y)=0$.

$\displaystyle \forall \varepsilon >0$, $\displaystyle \exists \delta (\varepsilon)$ such that if $\displaystyle |(x,y)-(x,x-1)|< \delta \Rightarrow |f(x,y)|< \varepsilon$.

$\displaystyle \Leftrightarrow |\sqrt{(y-x+1)^2}|<\delta \Rightarrow |y^2-(x-1)^2|<\varepsilon \Leftrightarrow y-x+1 < \delta \Rightarrow |y^2-(x-1)^2|< \varepsilon$ which work if I take $\displaystyle \delta=\sqrt \varepsilon$. Hmm no it doesn't work. I must sleep now, will think about it.

If you have any help to provide, it's always a pleasure for me. Don't hesitate.

Maybe you find it difficult to prove that $\displaystyle f(x,y)$ goes to $\displaystyle f(x_0,y_0)$ as $\displaystyle (x,y)\in A$ tends to $\displaystyle (x_0,y_0)\in A$. Initially I wouldn't have thought of even trying to prove this: one usually assumes that a polynomial function $\displaystyle f:\mathbb{R}^n\rightarrow \mathbb{R}$ of the coordinates of a point of $\displaystyle \mathbb{R}^n$ is continuous without further addo. This is because the projection of a point of $\displaystyle \mathbb{R}^n$ onto its k-th coordinate is continuous, and that sums and products of continous functions are continuous.

Thus there is no reason to prove continuity of a polynomial in the coordinates for each and every such polynomial using a pure delta-epsilon argument: it would be too much work (depending on the complexity of the polynomial) and it wouldn't be particularly enlightening either.