Continuity of a 2 variable function

**arbolis** · July 29th 2009, 08:55 PM

Let $A=\{ (x,y) \in \mathbb{R}^2 :y \leq x-1 \}$ and let $f(x,y)$ be defined in all $\mathbb{R}^2$ by $f(x,y)= \begin{cases} y^2-(x-1)^2 \hspace{2cm} \text{if (x,y)} \in A \\ 0 \hspace{3 cm} \text{if (x,y) is not in A} \end{cases}$ .
1)Analyze the continuity of $f$ in all $\mathbb{R}^2$ .
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 $y=x-1$ (a line) from the domain and send it to the image as the line $y=0$ (or curve?). I also realize that it sends all the elements under the curve $y=x-1$ 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 : $f$ is continuous in all $\mathbb{R}^2$ except maybe in $y=x-1$ . I'll try to prove it is :
I want to prove that $\lim _{y \to x-1} f(x,y)=0$ .
$\forall \varepsilon >0$ , $\exists \delta (\varepsilon)$ such that if $|(x,y)-(x,x-1)|< \delta \Rightarrow |f(x,y)|< \varepsilon$ .
$\Leftrightarrow |\sqrt{(y-x+1)^2}|<\delta \Rightarrow |y^2-(x-1)^2|<\varepsilon \Leftrightarrow y-x+1 < \delta \Rightarrow |y^2-(x-1)^2|< \varepsilon$ which work if I take $\delta=\sqrt \varepsilon$ . 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.

**Media_Man** · August 1st 2009, 06:38 AM

Let $|y-(x-1)|<\delta$

Now $|y^2-(x-1)^2|=|y-(x-1)||y+(x-1)|<$ $\delta|y-(x-1)+2(x-1)|\leq\delta(|y-(x-1)|+|2(x-1)|)<\delta(\delta+2|x-1|)$

So given an $\epsilon$ , $\delta$ must satisfy $\delta^2+2\delta|x-1|<\epsilon$ . Since it is dependent on x, this can never be achieved.

Visually speaking, the higher x goes, the steeper the graph becomes when crossing into A. As x goes to infinity, this steepness tends asymptotically to a $90^\circ$ fold in the graph. Therefore given a height variation $\pm\epsilon$ , there can never exist a circle in the xy-plane of any radius $\delta$ that can trap an area of the graph with no height variations less than $\pm\epsilon$ . You can always proceed Northeast along the line $y=x-1$ and find a spot a little bit steeper.

I know this doesn't count as a formal proof. What is the conventional way to disprove continuity via epsilon/delta?

**Failure** · August 1st 2009, 08:06 AM

Originally Posted by arbolis

Let $A=\{ (x,y) \in \mathbb{R}^2 :y \leq x-1 \}$ and let $f(x,y)$ be defined in all $\mathbb{R}^2$ by $f(x,y)= \begin{cases} y^2-(x-1)^2 \hspace{2cm} \text{if (x,y)} \in A \\ 0 \hspace{3 cm} \text{if (x,y) is not in A} \end{cases}$ .
1)Analyze the continuity of $f$ in all $\mathbb{R}^2$ .
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 $y=x-1$ (a line) from the domain and send it to the image as the line $y=0$ (or curve?). I also realize that it sends all the elements under the curve $y=x-1$ 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 : $f$ is continuous in all $\mathbb{R}^2$ except maybe in $y=x-1$ .

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 $y=x-1$ . f happens to be continuous at any inner point of either A or $\mathbb{R}^2\backslash A$ : 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 $y=x-1$ .

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 $y=x-1$ the way you do below:

I'll try to prove it is :
I want to prove that $\lim _{y \to x-1} f(x,y)=0$ .
$\forall \varepsilon >0$ , $\exists \delta (\varepsilon)$ such that if $|(x,y)-(x,x-1)|< \delta \Rightarrow |f(x,y)|< \varepsilon$ .
$\Leftrightarrow |\sqrt{(y-x+1)^2}|<\delta \Rightarrow |y^2-(x-1)^2|<\varepsilon \Leftrightarrow y-x+1 < \delta \Rightarrow |y^2-(x-1)^2|< \varepsilon$ which work if I take $\delta=\sqrt \varepsilon$ . 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 $f(x,y)$ goes to $f(x_0,y_0)$ as $(x,y)\in A$ tends to $(x_0,y_0)\in A$ . Initially I wouldn't have thought of even trying to prove this: one usually assumes that a polynomial function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ of the coordinates of a point of $\mathbb{R}^n$ is continuous without further addo. This is because the projection of a point of $\mathbb{R}^n$ 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.

**arbolis** · August 1st 2009, 10:46 AM

(Quoting does not work, sorry).
To Media_Man, ok so if I understand well $f$ is not continuous at $y=x-1$ hence not continuous in all $\mathbb{R}^2$ .
To Failure : ok, so if I understand well $f$ is continuous in all $\mathbb{R}^2$ .

Originally Posted by Media_Man

I know this doesn't count as a formal proof. What is the conventional way to disprove continuity via epsilon/delta?

with a 2 variables function, we generally approach the limit via different curves. If the limits are different, the real limit does not exist.

**Media_Man** · August 1st 2009, 11:04 AM

Allow a point $(x,y)$ to approach the line $y=x-1$ via the following curve: $y=\frac{x^2-2}{x+1}$ (whose slant asymptote is $y=x-1$ ). This curve is also completely in A. However, $\lim_{x\to\infty}f(x,y)=\lim_{x\to\infty}y^2-(x-1)^2=\lim_{x\to\infty}\left(\frac{x^2-2}{x+1}\right)^2-(x-1)^2=-4\neq 0$

This thread seems confusing to me, as visualizing this function, it seems perfectly continuous by the traditional (intuitive) definition that $\lim_{x\to a}f(x)=f(a)$ for all a.

**Failure** · August 1st 2009, 11:31 AM

Originally Posted by Media_Man

Allow a point $(x,y)$ to approach the line $y=x-1$ via the following curve: $y=\frac{x^2-2}{x+1}$ (whose slant asymptote is $y=x-1$ ). This curve is also completely in A. However, $\lim_{x\to\infty}f(x,y)=\lim_{x\to\infty}y^2-(x-1)^2=\lim_{x\to\infty}\left(\frac{x^2-2}{x+1}\right)^2-(x-1)^2=-4\neq 0$

This thread seems confusing to me, as visualizing this function, it seems perfectly continuous by the traditional (intuitive) definition that $\lim_{x\to a}f(x)=f(a)$ for all a.

Let it be noted that the definition of f being continuous at point a via the condition that $\lim_{x\to a}f(x)=f(a)$ holds is not at all merely "intuitive": it is perfectly valid.

The problem, as I see it, is that your method of "disproving continuity" of f does not work. To disprove continuity of f you must show that if $(x,y)$ approaches a specific point $(x_0,y_0)$ in a certain way, the values $f(x,y)$ do not approach $f(x_0,y_0)$ . Your letting x tend to infinity, while requiring that $y=\frac{x^2-2}{x+1}$ , is definitely not a case of approaching a specific point $(x_0,y_0)$ of the domain of f as required to refute continuity at $(x_0,y_0)$ .

**Failure** · August 1st 2009, 12:05 PM

To Failure : ok, so if I understand well $f$ is continuous in all $\mathbb{R}^2$ .

Well, yes: at least that's what I argued. The question is whether you agree with that argument or not.
Of course, you could also use an epsilon-delta style argument to show that $(x,y)\mapsto y^2-(x-1)^2$ is continuous at an arbitrary point $(x_0,y_0)$ . However, this requires a bit of grunt work. Just consider the following: assume $\varepsilon>0$ is given, the question is whether it is possible to find a $\delta >0$ such that $\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ implies that

$|f(x,y)-f(x_0,y_0)|=|(y^2-(x-1)^2)-(y_0^2-(x_0-1)^2)|<\varepsilon$

Using the fact that $a^2-a_0^2=(a-a_0)(a+a_0)$ and the triangle inequality it is possible to specify such a $\delta >0$ , but the whole thing is just way too complicated to write up (for an impatent fellow like me at least) since, as I wrote in my first post in this thread, all functions $f:\mathbb{R}^n\rightarrow \mathbb{R}$ whose function term is defined as a polynomial in the coordinates of its argument is continuous because: all projections $(x_1,\ldots,x_n)\mapsto x_k$ are continous and the product and the sum of continuous functions are continuous. (I forgot to mention the trivial case: the constant function must also be known to be continuous.) Every "polynomial f in the coordinates of its argument" can be gotten by adding and multiplying suitably chosen functions that are already known to be continuous (i.e. projections onto coordinate subspaces and the constant functions).

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