The Unapologetic Mathematician

Mathematics for the interested outsider

Cauchy’s Condition

We defined the real numbers to be a complete uniform space, meaning that limits of sequences are convergent if and only if they are Cauchy. Let’s write these two out in full:

  • A sequence a_n is convergent if there is some L so that for every \epsilon there is an N such that n>N implies |a_n-L|<\epsilon.
  • A sequence a_n is Cauchy if for every \epsilon there is an N such that m>N and n>N implies |a_m-a_n|<\epsilon.

See how similar the two definitions are. Convergent means that the points of the sequence are getting closer and closer to some fixed L. Cauchy means that the points of the sequence are getting closer to each other.

Now there’s no reason we can’t try the same thing when we’re taking the limit of a function at \infty. In fact, the definition of convergence of such a limit is already pretty close to the above definition. How can we translate the Cauchy condition? Simple. We just require that for every \epsilon>0 there exist some R so that for any two points x,y>R we have |f(x)-f(y)|<\epsilon.

So let’s consider a function f defined in the ray \left[a,\infty\right). If the limit \lim\limits_{x\rightarrow\infty}f(x) exists, with value L, then for every \epsilon>0 there is an R so that x>R implies |f(x)-L|<\frac{\epsilon}{2}. Then taking y>R as well, we see that

|f(x)-f(y)|=|(f(x)-L)-(f(y)-L)|\leq|f(x)-L|+|f(y)-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon

and so the Cauchy condition holds.

Now let’s assume that the Cauchy condition holds. Define the sequence a_n=f(a+n). This is now a Cauchy sequence, and so it converges to a limit A, which I assert is also the limit of f. Given an \epsilon>0, choose an R so that

  • |f(x)-f(y)|<\frac{\epsilon}{2} for any two points x and y above R
  • |a_n-A|<\frac{\epsilon}{2} whenever a+n\geq B

Just take a B for each condition, and go with the larger one. In fact, we may as well round B up so that B=a+N for some natural number N. Then for any b>B we have

|f(b)-A|=|(f(b)-f(B))+(f(B)-A)|\leq|f(b)-f(B)|+|a_N-A|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon

and so the limit at infinity exists.

In the particular case of an improper integral, we have I(b)=\int_a^bf\,d\alpha. Then I(c)-I(b)=\int_b^cf\,d\alpha. Our condition then reads:

For every \epsilon>0 there is a B so that c>b>B implies \left|\int_b^cfd\alpha\right|<\epsilon.

April 23, 2008 Posted by | Analysis, Calculus | 3 Comments

   

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