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Topic: Continuity and Differentiability (Read 1326 times) |
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Neelesh
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Continuity and Differentiability
« on: Sep 21st, 2005, 3:17am » |
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Can we have a function from R to R which is discontinuous at every point in the domain?
Can we have a function from R to R which is continuous everywhere but differentiable nowhere?
By function, I mean "total function" and not "partial function"
(Sorry if this is a repost. Could'nt find it though.)
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Barukh
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Re: Continuity and Differentiability
« Reply #1 on: Sep 21st, 2005, 5:04am » |
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The answers to both questions is "yes". These are classical problems, even if not on this site.
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Icarus
wu::riddles Moderator
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Re: Continuity and Differentiability
« Reply #2 on: Sep 21st, 2005, 2:55pm » |
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Not only do such functions exist, but they are both uniformly dense in their appropriate function spaces. I.e., you can uniformly approximate any function to any desired accuracy >0 with a function that is nowhere continuous, and you can uniformly approximate any continuous function with a continuous nowhere differentiable one.
For more fun, try to find a function that is continuous only for irrational values.
These problems have "sort of" been posted before, in that they are consequences of a problem posted not that long ago: determining what sets of real numbers can be the set of discontinuities, or of non-differentiability for some function.
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| « Last Edit: Sep 21st, 2005, 5:47pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Re: Continuity and Differentiability
« Reply #3 on: Sep 25th, 2005, 7:19pm » |
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Thinking about my previous reply, I've noticed that proving the first result I mentioned may be harder than proving the less obvious second result (at least, the second result is easier if you can take the Weierstrass Approximation Theorem as given). So I have decided to throw out this challenge:
Given an arbitrary function f: R --> R, and a value h > 0, show that there exists a function g: R --> R such that |g(x) - f(x)| < h for all x, and g is nowhere continuous.
At first thought, one would simply add a small nowhere continuous function to f in order to get g. But since you have no restrictions on f, there is a possibility that discontinuities in f could cancel those in the small function, to give a point of continuity for g.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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