Continuity of two-variable function

[twoflower]

Hi, I have some troubles understanding the basic facts about investigating the continuity of two-variable functions.
Our professor gave us very simple example to show us the basic facts:Very important is that projections are continuous, it means

<br /> \pi_1 :[x,y] \longmapsto x<br />

<br /> \pi_2 :[x,y] \longmapsto y<br />

are continuous.
Now let's have this function:

<br /> f(x,y) = \sqrt{x^2+y^2}<br />

It's continuous on the whole \mathbb{R}^2. Why?

1. Step:
Projections are continuous

2. Step:

<br /> x,y \longmapsto x^2 + y^2...continuous

3. Step:
Square root is continuous on [0,\infty)
What I don't fully understand is the second step. I can't see why, from the fact that projections are continuous, we can say that

<br /> x^2 + y^2<br />

is continous. Of course I didn't expect any other result, when you look at it it's obvious that it will be continuous, but I just can't see the rigorous mathematical background.

You know, I see it is some equivalent of the limit of product of single variable function (? is it the right expression ?), but I don't think it's sufficient. I would need some equivalent for two-variable functions...

Hope you understand my problem :)

Thank you very much.

 

[vanesch]

What I don't fully understand is the second step. I can't see why, from the fact that projections are continuous, we can say that
<br /> x^2 + y^2<br />
is continous. Of course I didn't expect any other result, when you look at it it's obvious that it will be continuous, but I just can't see the rigorous mathematical background.

I'll give it a try, but I'm not a mathematician, so I might get it wrong ; feel free to correct me if that's the case.
Consider the function a \rightarrow a^2. That's a polynomial and hence continuous.
Now, consider the first function:
(x,y) \rightarrow x \rightarrow x^2
This is a projection, followed by our squaring, so that's a continuous function from (x,y) to x^2.
Next, consider the second function:
(x,y) \rightarrow y \rightarrow y^2
This is also a projection, followed by squaring, so that's a continuous function too, from (x,y) to y^2.
The sum of two continuous functions, is continuous, so if we add our two functions, we have (x,y) \rightarrow x^2 + y^2
should be continous.
Did I sneak in something ?

 

[siddharth]

Hi, I have some troubles understanding the basic facts about investigating the continuity of two-variable functions.

Twoflower,

Let the point M_0 (x_0,y_0) belong to the domain of the function f(x,y). Then the function z=f(x,y) is continuous at the point M_0 (x_0,y_0) if we have

\lim_{\substack{x\rightarrow x_0\\y\rightarrow y_0}} f(x,y) = f(x_0,y_0)

and (x,y) approaches M_0 (x_0,y_0) in arbitary fashion all the while remaining in the domain of the function.

Now if x=x_0 + \Delta x , y = y_0 + \Delta y then the above condition can be rewritten as

\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} f(x_0 + \Delta x,y_0 + \Delta y) = f(x_0,y_0)

\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} [f(x_0 + \Delta x,y_0 + \Delta y) - f(x_0,y_0)] = 0

or
\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z = 0

So for z=\sqrt{x^2+y^2},
what is
\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z for any point in the domain?

 

[JasonRox]

Twoflower,

Let the point M_0 (x_0,y_0) belong to the domain of the function f(x,y). Then the function z=f(x,y) is continuous at the point M_0 (x_0,y_0) if we have

\lim_{\substack{x\rightarrow x_0\\y\rightarrow y_0}} f(x,y) = f(x_0,y_0)

and (x,y) approaches M_0 (x_0,y_0) in arbitary fashion all the while remaining in the domain of the function.

Now if x=x_0 + \Delta x , y = y_0 + \Delta y then the above condition can be rewritten as

\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} f(x_0 + \Delta x,y_0 + \Delta y) = f(x_0,y_0)

\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} [f(x_0 + \Delta x,y_0 + \Delta y) - f(x_0,y_0)] = 0

or
\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z = 0

So for z=\sqrt{x^2+y^2},
what is
\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z for any point in the domain?

Doesn't get any easier than this.

It's very simliar, if not the same, as continuity for a single variable.

 

[HallsofIvy]

Doesn't get any easier than this.

It's very simliar, if not the same, as continuity for a single variable.

Actually, it does- put it in polar coordinates and it IS continuity for a single variable!

The point made earlier, by vanesch, is that you cannot be certain of getting the correct limit of
\lim_{\substack{\Delta{x}\rightarrow 0\\\Delta{y}\rightarrow 0}} \Delta z

By simply taking \Delta y going to zero and then taking \Delta x going to 0. However, you can find the limit by seeing what happens when (\Delta x, \Delta y) is very close to (0,0). In polar coordinates that is measured by the single variable r. What do you get if you put this problem in polar coordinates?