Monsieur Ejemplé

Give an example of a function f:R^2 --> R s.t. the function is not continuous at zero, but for any (straight) line L passing through the origin, the induced function f|L is continuous on L.

>>1
Sorry, I don't know what those symbols mean. You need to provide a key to explain what they mean. Don't be such an idiot in the future.

>>2
If you're that stupid, it won't make a difference whether or not you know what the (self-evident) symbols mean.

f(x,y) = 1 when y = x^2 and x != 0
f(x,y) = 0 everywhere else.

There are more nicely defined functions with the same property, but I'm too lazy to come up with one right now.

>>4
Correct. A rather nice one is

f(x,y) = q when y = (p/q)x, x != 0
f(x,y) = 0 where y = rx, r irrational, x != 0
f(0,0) = 0

i only belief in science!!!!

>>4

I don't get it. Maybe I'm misenterpreting the induced function, but that way I see it any straight line through the origin would be 0, except at one point, where it would be one, except in the freak case the line is y=0, then the induced function would be 0 continously.

Am I not quite getting the idea of induced functions? Never met them

>>7

They're not very nice.

>>7
Note the "x != 0"; the function is 1 on the parabola EXCEPT on the origin, and 0 everywhere else.

>>9


So?

There's an intersection between the parabola and every line through the origin, except the trivial x-axis line. Therefore every line through the origin has one point where f(x,y)=1 and it is not continour at that point.

>>10
Oh, oops. I read the OP's request as "continuous at 0 on L", not continuous on the entirety of L. I think it can be rectified by redefining the function somewhat;

f(x, y) = 1 if y = x^2 and x != 0
        = 0 if y < x^2/2 or y > 3x^2/2 and x != 0
        = 0 if x = 0
        = 2y/x^2 - 1 if y >= x^2/2 and y < x^2
        = -2y/x^2 + 3 if y > x^2 and y <= 3x^2/2.

basically the same as before except it slopes up to the parabola. I know there's some rational function that has the property the OP is looking for that would be a lot nicer, but I can't fucking remember it.

OP
>>11
I wasn't thinking straight when I said your first one was correct.  Either way, my one in >>5 is pretty good.

>>10
You're right. The idea of an induced function is basically what it looks like.

>>12
Except that I also need f = 0 when x = 0

>>12
Your example in 5 has issues. For instance, on the line y = x. f(x,y) = 1 for x != 0 because y = 1/1 * x. But f(0,0) = 0, so it isn't continuous on the entirety of the line.

A nicer example still is:

  f(0,0) = 0
  f(x,y) = yx^2 / (x^4 + y^2) elsewhere

On any line y=mx we have f(x,y) = mx^3 / (x^4 + m^2x^2) = mx / (x^2 + m^2), and the limit as x->0 is 0, so the restriction of the function to the line y=mx is continuous everywhere.  On the line x=0 we have f(0, y) = 0 / y^2, and the limit as y->0 is 0, so again the function is continuous everywhere.

But if you consider the restriction of the function to the parabola y=x^2, you get f(x, x^2) = x^4 / 2x^4 = 1/2, so the function is not continuous at the origin.

There's an exercise about this in the Baby Rudin book.
(http://en.wikipedia.org/w/index.php?title=Walter_Rudin)

>>14

My bad, I meant to put f(x,y) = qx for y = (p/q)*x