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How do we know if a function is infinitely continuous? That mean it is continuous at every point?

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We prove it. What's the function? – Qiaochu Yuan May 10 '12 at 19:40
A function $f:A\to B$ is said to be continuous on $A$ if for every $\varepsilon>0$, there is a $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$ for all $x,a\in A$. – Josué Molina May 10 '12 at 19:42
If by "infinitely continuous" you are refering to the symbol $\mathcal{C}^{\infty}$, this means that at each point, the function has derivatives of all orders; in particular, it is continuous and differentiable everywhere, the derivative is continuous and differentiable at every point , the second derivative is continuous and differentiable at every point, etc. – Arturo Magidin May 10 '12 at 19:42
@QiaochuYuan - Is there any process or algorithm that is for every function on the proof? Or just we have to arrange the usage of theorem or order of theorem for different theorem? – Victor May 10 '12 at 19:44
@Victor: no. As with many problems, it's possible to encode undecidable problems (e.g. the Halting problem) into the problem of whether some arbitrary function is continuous. In practice these artificial examples don't come up and generally speaking you use simple established facts about continuous functions. – Qiaochu Yuan May 10 '12 at 19:49
up vote 2 down vote accepted

You shouldn't say "infinitely continuous" for this. A function is continuous if it is continuous at every point of its domain (that is the adopted definition in, say, real analysis).

Going with the definition of continuity of a function $f : D \to \mathbb R$ at a point $x_0$ is a starting point : $$ \forall \varepsilon > 0, \exists \delta > 0 \quad s.t. \forall x \in D, \quad |x-x_0| < \delta \quad \Rightarrow \quad |f(x)-f(x_0)| < \varepsilon $$ but in general one wants to use the fact that most elementary functions are continuous (polynomials, rational polynomials where the denominators don't vanish, fractional exponents, sines, cosines, exponentials, log, and sums/products/compositions of those) because working with the definition all the time will make you crazy. But it's good to be able to work with it though.

Hope that helps,

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Is it possible that he means absolutely continuous? That is a term used in probability used for distribution functions that have a derivative everywhere and hence a density function exists for it. – Michael Chernick May 10 '12 at 19:57
@Michael Chernick : He said "That means it is continuous at every point." I don't think he means absolutely continuous. Plus, if you check his profile, he's an high school student. Highly unlikely. – Patrick Da Silva May 10 '12 at 20:03
After I said it I had a feeling that the possiblity was very remote. Is infinitely continuous a real math term for everywhere continuous? – Michael Chernick May 10 '12 at 20:05
@Michael Chernick : Not that I heard of. It sounded like a term one would use when he does not know how to say it and it just gets out the way it feels natural to express it. – Patrick Da Silva May 10 '12 at 20:23
@Patrick: Of course. That doesn't change the fact that he did not actually make the assertion that you attributed to him. I wasn't objecting to your interpretation; I was merely pointing out that your justification to Michael Chernick is a bit specious. – Brian M. Scott May 10 '12 at 21:34

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