Arzela Ascoli Theorem Statement Proof Conditions and Examples of Equicontinuity and Compactness
Arzelà-Ascoli Theorem is used to determine the nature of the subsequence of every sequence of continuous functions. Here, we will first discuss the statement, followed by its proof and applications.
The Curve of Continuous function f
The above figure shows continuous functions f and subsequence $f_{m}$. So, we will be discussing the nature of the subsequence of $f_{m}$. In this article, we will also discuss the application of the theorem along with some of the limitations of the theorem. Some examples related to the theorem will also be solved here.
History of Mathematician
Two great mathematicians are associated with the Arzelà-Ascoli Theorem.
Cesare Arzelà
Name: Cesare Arzelà
Born: 6 March 1847
Died: 12 March 1912
Field: Mathematician
Nationality: Italian
Giulio Ascoli
Name: Giulio Ascoli
Born: 20 January 1843
Died: 12 July 1896
Field: Mathematics
Nationality: Italian
Statement of Arzelà-Ascoli Theorem
If a sequence, $\left\{f_{1}\right\}_{n=1}^{\infty}$ in $C(X)$ is bounded and equicontinuous, then it holds a uniformly convergent subsequence.
Mathematically it can be written as follows:
a) "F $\subset C(X)$ is bounded" means there exists a positive constant $M<\infty$ such that $|f(x)| \leq M$ for each $x \in X$ for all $f \in F$, and
(b) "F $\subset C(X)$ is equicontinuous" means for every $\varepsilon>0$ there exists $\delta>0$ such that for $x, y \in X:$
$d(x, y)<\delta \Rightarrow|f(x)-f(y)|<\varepsilon$ for all $f \in F$
Here, $d$ is the metric on $X$, and $\delta$ depends only on $\varepsilon$.
Proof of Arzelà-Ascoli Theorem
Firstly, we show that the compact metric space $X$ is separable, i.e., it has a finite dense subset $S$.
For a given positive integer $n$ and a point $x \in X$.
Let $B\left(x, \dfrac{1}{n}\right)=\{y \in X: d(x, y)<\dfrac{1}{n}\}$ be the open ball with radius $\dfrac{1}{n}$, centred at $x$.
For a given value of $n$, the set of all these balls as $x$ ranges through $X$ is an open cover of $x$ since $X$ is compact.
Thus, there is a countable subcollection that also covers $X$.
Let $S_{n}$ be the set of centres of the balls in the finite subcollection defined above.
Thus, $S_{n}$ is a finite subset of $X$, consequently " $\dfrac{1}{n}$-dense" in that each point of $X$ lies within $\dfrac{1}{n}$ of a point of the set $S n$.
Therefore, the union $S$ of all the sceneries $S_{n}$ is finite and dense in $X$.
Now, let us find a subsequence of $\left\{f_{n}\right\}$ that converges pointwise on $S$.
Let $\left\{x_{1}, x_{2}, \ldots\right\}$ be the elements of $S$.
As we know. then the numerical sequence $\left\{f_{n}\left(x_{1}\right)\right\}_{n=1}^{\infty}$ is bounded.
Thus, by Bolzano-Weierstrass, we can say that it has a convergent subsequence, and this can be written using double subscripts as:
$\left\{f_{1}, n\left(x_{1}\right)\right\}_{n=1}^{\infty}$
In the same way, we can show that the numerical sequence $\left\{f_{1}, n\left(x_{2}\right)\right\}_{n=1}^{\infty}$ is bounded, so it has a convergent subsequence $\left\{f_{2}, n\left(x_{2}\right)\right\}_{n=1^{\infty}}$.
From the above, we can say that the sequence of functions $\left\{f_{2}, n\right\}_{n=1^{\infty}}$, since it is a subsequence of $\left\{f_{1}, n\right\}_{n=1}^{\infty}$, converges at both $x_{1}$ and $x_{2}$.
Moving in this technique, we obtain a finite collection of sub-sequences of the initial sequence, as:
$f_{1,1} f_{1,2} f_{1,3} \cdots f_{2,1} f_{2,2} f_{2,3} \cdots f_{3,1} f_{3,2} f_{3,3} \cdots \quad f_{n, 1} f_{n, 2} f_{n, 3} \cdots$
Here, the sequence of the $n$-th row converges at $x_{1}, \ldots, x_{n^{\prime}}$ and every row is a subsequence of its previous row.
Therefore, the diagonal sequence in the above, i.e., $\left\{f_{n, n}\right\}$ is a subsequence of the initial sequence $\left\{f_{n}\right\}$ that converges at each point of $S$.
Let $\left\{g_{n}\right\}$ be the diagonal subsequence that is convergent at each point of the dense set $S$ that is created from the overhead step.
Let $\varepsilon>0$ and $\delta>0$ then by equicontinuity of the original sequence, so that $d(x, y)<\delta$ which implies $\left|g_{n}(x)-g_{n}(y)\right|<\dfrac{\varepsilon}{3}$ for each $x, y \in x$ and $n$ is a positive integer.
Specify $M>\dfrac{1}{\delta}$ so that the countable subset $S M \subset S$ that we produced in the initial steps is $\delta$-dense in $X$.
As $\left\{g_{n}\right\}$ converges at each point of $S M$, there exists $N>0$ such that; $n, m>N \Rightarrow\left|g_{n}(s)-g_{m}(s)\right|<\dfrac{\varepsilon}{3}$ for all $s \in S M$.
Consider $x \in X$ then $x$ lies within $\delta$ of some $s \in S M$, so if $n, m>M$ :
$\left|g_{n}(x)-g_{m}(x)\right| \leq\left|g_{n}(x)-g_{n}(s)\right|+\left|g_{n}(s)-g_{m}(s)\right|+\left|g_{m}(s)-g_{m}(x)\right|$
In $R H S$, the first and last terms are $<\dfrac{\varepsilon}{3}$ by our preference of $\delta$.
Thus, for a given $\varepsilon>0$ we can produce $N$ such that for each $x \in X, m, n>N \Rightarrow\left|g_{n}(x)-g_{m}(x)\right|<\dfrac{\varepsilon}{3}+\dfrac{\varepsilon}{3}+\dfrac{\varepsilon}{3}=\varepsilon$.
Therefore, on $X$ the subsequence $\left\{g_{n}\right\}$ of $\left\{f_{n}\right\}$ is uniformly Cauchy, and uniformly convergent.
Hence the proof of the theorem.
Limitations of Arzelà-Ascoli Theorem
Arzelà-Ascoli Theorem is quite complex to understand as it requires in-depth knowledge of the subject to understand the theorem.
Besides a wide range of applications, there are many proofs of the theorem for different branches of mathematics, such as metrics spaces, real analysis, topology, etc., which makes it confusing for readers to differentiate between them.
Applications of Arzelà-Ascoli Theorem
Applications to Functional Analysis in proving compactness of equicontinuous family of functions.
Applications to Ordinary Differential Equations in proving Peano’s Existence Theorem.
Applications to Complex Analysis in Riemann’s Mapping Theory.
Solved Examples
1. What are equicontinuous functions as used in the Statement of Arzelà-Ascoli Theorem?
Ans: In mathematical analysis, a family of functions is said to be equicontinuous if all the functions are continuous and they have equal variation in the given neighbourhood of the family of functions. In particular, the concept applies to countable families and thus sequences of functions.
Important Points/Formulas to Remember
For the subsequence of any function to be convergent, the sequence of uniform functions must be defined in closed and bounded intervals.
An interval is said to be bounded if there exists an upper and lower bound for the interval.
If the value of a function coincides with a common value at all values for a given interval, then the function is said to be convergent.\
Conclusion
The Arzelà-Ascoli theorem allows us to study compact sets in function spaces in Mathematical Analysis. This theorem shortens the checking of compactness for subsets of spaces of continuous function. Moreover, a lot of the topological spaces used in Real Analysis, Complex Analysis, and Functional Analysis are spaces of functions.
FAQs on Arzela Ascoli Theorem in Real Analysis and Functional Analysis
1. What is the Arzelà–Ascoli Theorem?
The Arzelà–Ascoli Theorem states that a family of real-valued continuous functions on a compact set has a uniformly convergent subsequence if it is equicontinuous and uniformly bounded. In its classical form for functions on a compact interval [a, b]:
- If a sequence (f_n) ⊂ C([a, b]) is uniformly bounded,
- And equicontinuous on [a, b],
- Then there exists a subsequence that converges uniformly to a continuous function.
2. What are the conditions of the Arzelà–Ascoli Theorem?
The key conditions of the Arzelà–Ascoli Theorem are uniform boundedness and equicontinuity on a compact domain. Specifically:
- Uniform boundedness: There exists M > 0 such that |f(x)| ≤ M for all functions f in the family and all x in the domain.
- Equicontinuity: For every ε > 0, there exists δ > 0 such that |x − y| < δ implies |f(x) − f(y)| < ε for all functions f in the family.
- The domain must be compact (e.g., a closed interval [a, b]).
3. What does equicontinuity mean in the Arzelà–Ascoli Theorem?
In the Arzelà–Ascoli Theorem, equicontinuity means all functions in the family share the same continuity control. Formally:
- For every ε > 0, there exists δ > 0,
- Such that for all functions f in the family,
- |x − y| < δ implies |f(x) − f(y)| < ε.
4. What does uniformly bounded mean in the Arzelà–Ascoli Theorem?
A family of functions is uniformly bounded if there exists a constant M such that |f(x)| ≤ M for every function f in the family and every x in the domain. This means:
- All graphs lie within the horizontal strip −M ≤ y ≤ M.
- The bound M is the same for all functions.
5. Why is compactness important in the Arzelà–Ascoli Theorem?
Compactness of the domain is essential because the theorem guarantees subsequential uniform convergence only on compact sets. For example:
- Closed and bounded intervals like [a, b] in ℝ are compact.
- On non-compact domains (e.g., ℝ), uniform convergence may fail even if functions are bounded and equicontinuous.
6. Can you give an example of the Arzelà–Ascoli Theorem?
A standard example is the sequence f_n(x) = sin(nx)/n on [0, 2π], which satisfies the conditions of the Arzelà–Ascoli Theorem. Check the conditions:
- Uniformly bounded: |sin(nx)/n| ≤ 1/n ≤ 1.
- Equicontinuous: Derivatives are bounded by 1, so oscillations are controlled.
7. What is the conclusion of the Arzelà–Ascoli Theorem?
The conclusion of the Arzelà–Ascoli Theorem is that there exists a uniformly convergent subsequence whose limit is continuous. More precisely:
- Given a uniformly bounded and equicontinuous sequence in C([a, b]),
- There exists a subsequence (f_{n_k}),
- Such that f_{n_k} → f uniformly on [a, b],
- Where f ∈ C([a, b]).
8. How is the Arzelà–Ascoli Theorem related to compactness in C([a, b])?
The Arzelà–Ascoli Theorem characterizes relatively compact subsets of C([a, b]) with the sup norm. Specifically:
- A subset of C([a, b]) is relatively compact,
- If and only if it is uniformly bounded and equicontinuous.
9. What is the difference between pointwise and uniform convergence in Arzelà–Ascoli?
The key difference is that uniform convergence controls convergence uniformly over the whole domain, while pointwise convergence does not. In detail:
- Pointwise convergence: For each fixed x, f_n(x) → f(x).
- Uniform convergence: sup|f_n(x) − f(x)| → 0 over the domain.
10. How do you prove the Arzelà–Ascoli Theorem (idea of proof)?
The proof of the Arzelà–Ascoli Theorem uses a diagonal argument and compactness of the domain. The main steps are:
- Choose a dense countable subset of the compact interval (e.g., rational points).
- Use boundedness to extract convergent subsequences at each point.
- Apply a diagonal argument to get pointwise convergence on the dense set.
- Use equicontinuity to upgrade pointwise convergence to uniform convergence.
