Prove that the space C[0,1] of continuous functions with the sup norm is complete.
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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∥∞, 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, {fn}, in C[0,1]. By the definition of a Cauchy sequence with respect to the sup norm, for any given ϵ>0, there exists an integer N such that for all m,n≥N, the sup norm of the difference between fm and fn is less than ϵ. This means that for any x∈[0,1], ∣fm(x)−fn(x)∣<ϵ.
Step 3: Pointwise Convergence
Since ∣fm(x)−fn(x)∣<ϵ for all x∈[0,1], the sequence of real numbers {fn(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, fn(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 ϵ>0, there exists N such that for m,n≥N, ∣fm(x)−fn(x)∣<ϵ for all x∈[0,1]. If we fix n≥N and let m→∞, then fm(x)→f(x). By the property of limits, ∣f(x)−fn(x)∣≤ϵ. This inequality holds for all x∈[0,1], which means that supx∈[0,1]∣f(x)−fn(x)∣≤ϵ. This shows that fn 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 fn is continuous on [0,1] and the sequence {fn} converges uniformly to f on [0,1], it follows that the limit function f must also be continuous on [0,1]. Therefore, f∈C[0,1].
Step 6: Conclusion
We have shown that any Cauchy sequence {fn} 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],∥⋅∥∞).