Suppose the restrictions of f to every straight line passing through a point a of R^n are continuous. Does this imply f is continuous at a?
- No. To see this, imagine a series of continuous functions which have a peak in the middle. Imagine this peak getting narrower and narrower. For a given ε, for each function you can still find an appropriate δ, but the δ gets smaller and smaller as you go (with limit 0) - so you can't find a single δ that will work for all of them.
Now what we have to do is find a sequence like that, and wrap them around the given point so that as you approach one particular angle, the δ required tends to 0 (on that angle, you just define the function to be constant).
For example, consider the function f(x, y) = e^(-k(x^2+y^2)). At the origin this is of course 1 for every k, but for a given ε we must be within √(ln (1-ε)/k) of the origin. So for any δ we choose, there is a k large enough to make f(x, y) more than ε distant from 1.
So if we let k = y/x, and hence
f(x, y) = e^((-y/x)(x^2+y^2)), x ≠ 0;
f(x, y) = 1, x = 0
then in any δ-circle we can find an x and y such that ||(x, y)|| < δ but ||1-f(x, y)|| > ε. So f(x, y) is not continuous at 0. But on any straight line we have either y = kx and hence f(x, y) = e^-k(x^2+y^2), which is continuous, or x = 0 and hence f(x, y) = 1.
A similar style of argument should work for the second question. If you have directional derivatives which increase in magnitude without limit as you approach one particular direction, and some finite directional derivative in that direction, then f will not be differentiable. I haven't constructed an actual example for this, though.
A saddle point is a critical point of f where some directions have a minimum and others have a maximum.
2007-04-12 17:51:16
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answer #1
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answered by Scarlet Manuka 7
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