MHB Proving Relatively Compactness in C([a, b]) using Arzelá-Ascoli Theorem

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SUMMARY

The discussion centers on proving that a set \( M \subset C([a, b]) \) is relatively compact using the Arzelá-Ascoli theorem. It is established that \( M \) is equicontinuous due to the condition \( |f(x)-f(y)| \leq L |x-y| \) for all \( f \in M \) and \( x, y \in [a, b] \). Additionally, \( M \) is uniformly bounded since \( |f(x_0)| \leq m \) for all \( f \in M \). Therefore, by the Arzelá-Ascoli theorem, \( M \) is relatively compact in \( C([a, b]) \).

PREREQUISITES
  • Understanding of the Arzelá-Ascoli theorem
  • Knowledge of equicontinuity in function spaces
  • Familiarity with uniform boundedness concepts
  • Basic principles of functional analysis, particularly in \( C([a, b]) \)
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  • Study the proof of the Arzelá-Ascoli theorem in detail
  • Explore examples of equicontinuous and uniformly bounded sets in \( C([a, b]) \)
  • Learn about compactness in functional spaces
  • Investigate applications of the Arzelá-Ascoli theorem in real analysis
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Mathematicians, particularly those specializing in functional analysis, students studying real analysis, and anyone interested in the properties of function spaces.

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I don't know how to solve this proof

Prove that a set $M \subset C([a, b])$ for which there exist $m. L> 0$ and $x_0 \in [a; b]$ such that $|f(x_0)| \leq{} m$ for all $f \in M$ and $|f(x)-f(y)| \leq{} L |x-y|$ for all $f\in M$ and for all $x,y \in [a,b]$ is relatively compact in $C([a, b])$

My attemptSince $L >0$ and $ f \in M $ we can conclude that M it is equicontinuous and we have that $ |f(x)-f(y)| \leq{} L |x-y| $and $f(x)$ is continuous and $[a,b]$ compact.
By Arzela-Acoli Theorem Then is relatively compact in$ C([a, b]$
How can I apply the Arzelá-Ascoli theorem?
Thanks

 
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The Arzelà-Ascoli theorem asserts that $M$ is relatively compact in $C[a,b]$ if $M$ is equicontinuous and pointwise bounded.

To see that $M$ is equicontinuous, fix positive number $\varepsilon$ and set $\delta = \varepsilon/L$. For all $x,y\in[a,b]$, $|x-y| < \delta$ and $f\in M$ implies $|f(x)-f(y)| \le L|x-y| < L(\varepsilon/L) = \varepsilon$. This shows, in fact, that $M$ is uniformly equicontinuous.

As for pointwise boundedness, fix an $x\in [a,b]$. For all $f\in M$, $|f(x)| \le |f(x_0)| + |f(x)-f(x_0)| \le m + L|x-x_0|$. Thus $\sup\limits_{f\in M} |f(x)| \le m + L|x-x_0|$.
 
for your question! You are on the right track with your attempt. Here is how you can apply the Arzelá-Ascoli theorem to solve this proof:

First, recall that the Arzelá-Ascoli theorem states that a subset $M \subset C([a, b])$ is relatively compact if and only if it is equicontinuous and uniformly bounded.

In this case, we have already established that $M$ is equicontinuous, since $|f(x)-f(y)| \leq{} L |x-y|$ for all $f\in M$ and for all $x,y \in [a,b]$.

Next, we need to show that $M$ is uniformly bounded. We know that there exists $m>0$ such that $|f(x_0)| \leq{} m$ for all $f \in M$. This means that all the functions in $M$ have a maximum absolute value of $m$. Therefore, $M$ is uniformly bounded by $m$.

Since $M$ is both equicontinuous and uniformly bounded, it follows that $M$ is relatively compact in $C([a, b])$ by the Arzelá-Ascoli theorem. This completes the proof.
 

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