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Why can't linear paths prove continuity in R^n?

  1. Oct 23, 2011 #1
    I know its a pretty elementary question, but I never felt like I've had any sort of reasonable explanation of why. As I understand, we can define continuity for a function f: ℝn→ℝ as:

    For any ε>0 there exists a δ>0 such that for all x st 0< lx - al < δ then lf(x) - f(a)l < ε

    Alright, so in other words, we can always define a "disc" in ℝn such that all function values belonging to the points in that disc are less that our chosen epsilon.

    So let's say we find that every linear path that passes through a is continuous at the point a. Let's also say that for each linear path, there is a maximum delta, call it δ', such that for all linear paths, for every ε we can choose δ' such that for all x 0< lx - al < δ' → lf(x) - f(a)l < ε .

    Doesn't this prove the continuity? I know there are counter examples that will show that if you take some other path, you may get a different limit. But it still seems to me that that the above should be true.
     
  2. jcsd
  3. Oct 23, 2011 #2

    Bacle2

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    If I understand you well, f(x,y)=xy/(x2+y2), with f(0,0)=0 is an exception. As a function of x alone, it is continuous; same as a function of yalone, but it is not continuous as a function of both (x,y); take, e.g. the limit along x=y, as x,y→0, which is not 0. Basically, a function into a product space is continuous if the restriction to each of the factor is continuous, but not the other way around, i.e., a map. from a product space into ℝ is not necessarily continuous when each of the component functions is continuous.
     
  4. Oct 23, 2011 #3
    What I really mean to ask is let's take all linear paths - i.e. with a function f:R^2->R take the limit along y=cx or x=cy for all values c, and if all these values converge to the same limit for which the function is defined, f(a,b), why doesn't that prove continuity at that point?

    To try and expand on why I have trouble with the idea that this does not prove continuity, I imagine a ball B(a,r) around a. Now, our definition of continuity says that for any value epsilon we can pick an r such that for any x ∈ B(a,r), lf(x)-F(a)l < ε. My thoughts are: how come the union of all sets of linear paths doesn't form a partition of this ball, hence imply continuity if all linear paths are continuous.
     
  5. Oct 23, 2011 #4
    Consider the following function:

    [tex]f(x) = \begin{cases} 0 & \mbox{if } x=0 \\ \frac{y^2}{x} & \mbox{if } x \neq 0 \end{cases}[/tex]

    Along the vertical line passing through zero, f is just the zero function, and so continuous. Along the line y=mx, f(x, mx) = mx2/x = mx at every point, and so is continuous. But f is not continuous, for if we approach along the path x(t) = t2, y(t) = t, we have f(t2, t) = t2/t2 = 1 where t≠0, and f(0, 0) = 0, so [itex]\lim_{t \rightarrow 0} f(t^2, t) = 1 \neq f(0, 0)[/itex]. Hence f is a function with a discontinuity at zero which is continuous along every line approaching zero.
     
    Last edited: Oct 24, 2011
  6. Oct 23, 2011 #5

    Bacle2

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    I think the more general answer to give is that, in this context, f is continuous if (iff, actually)
    xn→x implies f(xn)→f(x). In your case, you are considering convergence only along linear paths where xn→x.
     
    Last edited: Oct 23, 2011
  7. Oct 24, 2011 #6

    lurflurf

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    Say we want to show f is continuous at x so
    given epsilon>0 there exist delta(epsilon)>0 such that
    whenever |h|<delta(epsilon)
    we have
    |f(x+h)-f(x)|<epsilon

    now consider the function
    g(|h|,h/|h|)=f(x+h)-f(x);0 when h=0

    we want to show f is continuous at x so
    given epsilon>0 there exist delta(epsilon)>0 such that
    whenever |h|<delta(epsilon)
    we have
    |g(|h|,h/|h|)|<epsilon

    your linear paths tell us g is continuous with respect to |h| so
    given epsilon>0 there exist delta(epsilon)>0 such that
    whenever |h|<delta(epsilon,h/|h|)
    we have
    |g(|h|,h/|h|)|<epsilon

    the trouble is our epsilon depends upon h/|h| a dependency we cannot eliminate

    or just imagine a shrinking bag full of mayonnaise with a protrusion
     
  8. Oct 25, 2011 #7

    Bacle2

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    Maybe you can also see the issue of the linear path as saying that the restriction of the function to the reals ( a line being the reals) along all possible directions is continuous. But as Citan showed, the restriction to the reals being continuous is not enough; maybe you can say that these "samples" of the functions along lines do not give an accurate-enough description of the content of the function. It is a good question; I don't know if this helps answer.
     
  9. Oct 26, 2011 #8
    There are more intuitive examples. Just take a function on R^2 that is equal to zero everywhere, except along some curve (say, a parabola) approaching the origin. So, the graph is flat, except for one big infinitely thin ridge that is coming in. Like a piece of paper sticking up from the ground. The limit along linear paths will always be zero. But if you walk along the piece of paper to the origin, you get a different limit. Run your purported proof through on that example and you should really grasp what goes wrong.

    If you want an example that is continuous everywhere except the origin, you can imagine smoothing out my example appropriately.
     
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