We begin by taking an arbitrary Cauchy sequence of functions, f_n, in C[0,1]. By the definition of a Cauchy sequence with respect to the sup norm, for any given > 0, there exists an integer N such that for all m, n ≥ N, the sup norm of the difference between f_m and f_n is less than. This means that for any x [0,1], |f_m(x) - f_n(x)| <.

We begin by taking an arbitrary Cauchy sequence of functions, f_n, in C[0,1]. By the definition of a Cauchy sequence with respect to the sup norm, for any given > 0, there exists an integer N such that for all m, n ≥ N, the sup norm of the difference between f_m and f_n is less than. This means that for any x [0,1], |f_m(x) - f_n(x)| <.

Since |f_m(x) - f_n(x)| < for all x [0,1], the sequence of real numbers f_n(x) is a Cauchy sequence in R for each fixed x. Because the real number system R is complete, every Cauchy sequence in R converges to a limit in R. Thus, for each x, f_n(x) converges to some real number, which we define as f(x). This defines a pointwise limit function f on [0,1].

Since |f_m(x) - f_n(x)| < for all x [0,1], the sequence of real numbers f_n(x) is a Cauchy sequence in R for each fixed x. Because the real number system R is complete, every Cauchy sequence in R converges to a limit in R. Thus, for each x, f_n(x) converges to some real number, which we define as f(x). This defines a pointwise limit function f on [0,1].

From the Cauchy property, for any > 0, there exists N such that for m, n ≥ N, |f_m(x) - f_n(x)| < for all x [0,1]. If we fix n ≥ N and let m → ∞, then f_m(x) → f(x). By the property of limits, |f(x) - f_n(x)| ≤. This inequality holds for all x [0,1], which means that _x [0,1] |f(x) - f_n(x)| ≤. This shows that f_n converges uniformly to f.

From the Cauchy property, for any > 0, there exists N such that for m, n ≥ N, |f_m(x) - f_n(x)| < for all x [0,1]. If we fix n ≥ N and let m → ∞, then f_m(x) → f(x). By the property of limits, |f(x) - f_n(x)| ≤. This inequality holds for all x [0,1], which means that _x [0,1] |f(x) - f_n(x)| ≤. This shows that f_n converges uniformly to f.

Prove that the space C[0,1] of continuous functions with the sup norm is complete.

Answer: The space $C[0{,}1]$ of continuous functions with the sup norm is complete.

Step-by-step solution

Step 1: Define Completeness and Sup Norm

The space $C[0{,}1]$ consists of all continuous real-valued functions defined on the closed interval $[0{,}1]$ . The sup norm, denoted as $\|f\|_{\infty}$ , for a function $f$ in this space is defined as the supremum of the absolute values of $f(x)$ over the interval $[0{,}1]$ . A metric space is complete if every Cauchy sequence in the space converges to a limit that is also in the space.

Step 2: Consider a Cauchy Sequence

We begin by taking an arbitrary Cauchy sequence of functions, $\{f_n\}$ , in $C[0{,}1]$ . By the definition of a Cauchy sequence with respect to the sup norm, for any given $\epsilon > 0$ , there exists an integer $N$ such that for all $m, n \ge N$ , the sup norm of the difference between $f_m$ and $f_n$ is less than $\epsilon$ . This means that for any $x \in [0{,}1]$ , $|f_m(x) - f_n(x)| < \epsilon$ .

Step 3: Pointwise Convergence

Since $|f_m(x) - f_n(x)| < \epsilon$ for all $x \in [0{,}1]$ , the sequence of real numbers $\{f_n(x)\}$ is a Cauchy sequence in $\mathbb{R}$ for each fixed $x$ . Because the real number system $\mathbb{R}$ is complete, every Cauchy sequence in $\mathbb{R}$ converges to a limit in $\mathbb{R}$ . Thus, for each $x$ , $f_n(x)$ converges to some real number, which we define as $f(x)$ . This defines a pointwise limit function $f$ on $[0{,}1]$ .

Step 4: Uniform Convergence

From the Cauchy property, for any $\epsilon > 0$ , there exists $N$ such that for $m, n \ge N$ , $|f_m(x) - f_n(x)| < \epsilon$ for all $x \in [0{,}1]$ . If we fix $n \ge N$ and let $m \to \infty$ , then $f_m(x) \to f(x)$ . By the property of limits, $|f(x) - f_n(x)| \le \epsilon$ . This inequality holds for all $x \in [0{,}1]$ , which means that $\sup_{x \in [0{,}1]} |f(x) - f_n(x)| \le \epsilon$ . This shows that $f_n$ converges uniformly to $f$ .

Step 5: Continuity of the Limit Function

A fundamental theorem in analysis states that the uniform limit of a sequence of continuous functions is continuous. Since each $f_n$ is continuous on $[0{,}1]$ and the sequence $\{f_n\}$ converges uniformly to $f$ on $[0{,}1]$ , it follows that the limit function $f$ must also be continuous on $[0{,}1]$ . Therefore, $f \in C[0{,}1]$ .

Step 6: Conclusion

We have shown that any Cauchy sequence $\{f_n\}$ in $C[0{,}1]$ converges uniformly to a function $f$ , and this limit function $f$ is continuous. Since $f$ is continuous on $[0{,}1]$ , it belongs to the space $C[0{,}1]$ . This satisfies the definition of completeness for the metric space $(C[0{,}1], \|\cdot\|_{\infty})$ .

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