Cauchy

Open any advanced calculus text or work in any field of analysis, you are likely to relish in the glorious ideas of Agustin-Louis Cauchy and not even know it. Cauchy was born in Paris, France in 1789. He began his professional life as an engineer and progressively influenced mathematics by way of research, professorships (when he was not refusing compliance with local regulations) and dabbling in physics (as the chair of several physics departments).

Today, we present a proof of the Cauchy Criterion for summations, series, and, finally, uniform convergence.

We proceed as follows:

1) A sequence { $x_n$ } is a Cauchy sequence ${\longleftrightarrow}$ it is convergent.

Proof: Suppose that { $x_n$ } is Cauchy. By definition, we have:

$\forall \epsilon>0 \exists N \in \mathbb{N}| n,m>N \rightarrow |x_n -x_m| \leq \epsilon$ .

By our definition, it follows that for some $m=N$ , $|x_n|\leq |x_N|+\epsilon$ . Pick $N=1,2,3,4,...n$ so that we select ${n}$ so that:

$|x_{n}|\leq max(|x_{1}|,|x_{2}|,|x_{3}|,\dots,|x_n|)$

If we let M= ${max(|x_{1}|,|x_{2}|,|x_{3}|},\dots,|x_n|)$ , then it follows that $|x_n|\leq M$ . So, we say that { $x_n$ } is bounded.

Since { $x_n$ } is bounded, by the Bolzano-Wierstrauss Property, it follows that:

$\forall \epsilon>0 \exists K,N \in \mathbb{N}| n,K \ge N \rightarrow |x_{n_k} -L| \leq \frac{\epsilon}{2}$ .

And, as stated, we take our definition for { $x_n$ },

$\forall \epsilon>0 \exists N \in \mathbb{N}| n,m>N \rightarrow |x_n -x_m| \leq \frac{ \epsilon}{2}$ .

Now, suppose that $n \ge$ ; set $\forall{m,n}, m=n_{k}$ so that $k \ge K$ where,

$|x_{n}-L| \leq |x_n-x_{k_n}|+|x_{n_k}-L| \leq \epsilon$

This establishes that fact that { $x_n$ } is convergent.

Now, we prove that if { $x_n$ } is convergent, then it is a Cauchy sequence.

The proof involving the conditional that a sequence { $x_n$ } convergent, is Cauchy is fairly simple. Consider a sequence $x_k \rightarrow L$ for some $k>N% we merely apply the definition of convergence using$ latex {x_n}$ in our inequality so as to produce the desired result -being careful to consider the correct use of inequalities.

Now, consider how this may apply to summations. One of the conveniences of summations is that they basically act as a sequence in disguise -the previous criterion can be used with a few minor additions to prove the criterion for summations. However, we will act more formally.

The series

$\sum_{k=1 }^{\infty} a_{k}$

is said to be satisfy the necessary and sufficient conditions of the Cauchy criterion provided that:

$\forall \epsilon>0\quad\exists{N}\in\mathbb{N}\big{|}\quad N\rightarrow |\sum_{k=n}^{\infty}|\leq\epsilon,\quad \forall {N} \leq n<m< \infty$

Now, the criterion holds: a series $\sum_{k=1 }^{\infty} a_{k}$ converges $\longleftrightarrow$ the criterion holds. So, to formalize it, we would take the proofs from above and equate it to $\sum_{k=1 }^{\infty} a_{k}$ as:

$|x_n-x_m|\leq |\sum_{k=n}^{m} a_k|<\epsilon$

Everything else falls out given that if we move in the other direction of the biconditional, we get a convergent subsequence which by definition fulfills the criterion.

Moving on, we are not at the end of our post. We have only one more fact to prove -uniform convergence. Uniform convergence is the idea that a sequence of function converges to some limiting function (or just, “a function”) if the speed of that convergence for successive n’s is sufficiently fast to arrive at the limiting function in question. For example, lets consider the following function, $f(x)=x^n, \quad n\in\mathbb{N}$ and make that a sequence -denoted as:

$f_n{(x)}=x^n$

The basic idea, here, is that as we change n, we want to see if we can find some large value of n so that we have some value of our sequence that arrives at some function that is within all chose values in the range. Let’s show different values of the previously mentioned function and see what its behavior is.

The above image should help you see what is happening here. The “speed” at which we are converging to some limit value is not sufficient. That is, as we push out n, we see that there is always a large n so that our values are the same for all f’s. Also, even if I push the value of n out to 1,000 or 100,000 I still do not arrive at the value required to establish uniform continuity.

So, what is our definition for something to be called uniformly convergent? Let’s take a look:

Defintion: Let $\{f_n\}$ be a sequence of functions defined on a common domain $\mathbb{D}$ . $\{f_n\}$ converges uniformly to a function ${f}$ on $\mathbb{D}$ if $\forall \epsilon>0 \exists N \in \mathbb{N}$ such that

$\forall n>N, x\in\mathbb{D}\quad|f_n{(x)}-f(x)|<\epsilon$

What does this mean for applying the Cauchy Criterion. Well, two things: (1) we are dealing with sequences that are involved in a limiting process, (2) the idea of functions converging should, depending on their nature, “act” the same way as sequences or series. Let’s move on to the meat-and-potatoes.

Cauchy Criterion (defintion): Let $\{f_n\}$ be a sequence of functions defined on a common domain $\mathbb{D}$ . The sequence $\{f_n\}$ is said to be uniformly Cauchy on $\mathbb{D}$ if $\forall \epsilon>0 \exists N$ such that if ${n}\geq{N}$ and ${m}\geq{N}$ then it follows that $|f_n{(x)}-f(x)|< \epsilon$

Criterion: $\{f_n\}\rightarrow{f}$ on $\mathbb{D}$ IFF $\{f_n\}$ is uniformly Cauchy.

Proof: $\{f_n\}$ be Cauchy, we show that ${\longleftrightarrow}$ it is convergent. Suppose that $\{f_n\}$ is Cauchy. By definition, we have:

$\forall \epsilon>0 \exists N$ such that if ${n}\geq{N}$ and ${m}\geq{N}$ then it follows that $|f_n{(x)}-f(x)|< \epsilon$

By our definition, it follows that for some $m=N$ , $|f_n|\leq |f_N|+\epsilon$ . Pick $N=1,2,3,4,...n$ so that we select ${n}$ so that:

$|f_{n}|\leq max(|f_{1}|,|f_{2}|,|f_{3}|,\dots,|f_n|)$

If we let M= ${max(|f_{1}|,|f_{2}|,|f_{3}|},\dots,|f_n|)$ , then it follows that $|x_n|\leq M$ . So, we say that { $f_n$ } is bounded.

Since { $f_n$ } is bounded, by the Bolzano-Wierstrauss Property, it follows that:

$\forall \epsilon>0 \exists K, N \in \mathbb{N}| n,K \ge N \rightarrow |f_{n_k} -f_{1}| \leq \frac{\epsilon}{2}$ .

And, as stated, we take our definition for { $fx_n$ },

$\forall \epsilon>0 \exists N \in \mathbb{N}| n,m>N \rightarrow |f_n -f_m| \leq \frac{ \epsilon}{2}$ .

Now, suppose that $n \ge$ ; set $\forall{m,n}, m=n_{k}$ so that $k \ge K$ where,

$|f_{n}-f_{1}| \leq |f_n-f_{k_n}|+|f_{n_k}-f_{1}| \leq \epsilon$

This establishes that fact that { $f_n$ } is uniformly convergent.

The other conditional is easy. If it is convergent, then by definition, it is Cauchy.

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John Piccolo

"…..no firm demonstration can be made from the success of hypotheses." -G.W. Leibniz

Cauchy

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