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Chapter 7: Problem 18

Let \(f\) be continuous on \([a, b]\), let \(f(x) \geq 0\) for \(x \in[a, b]\), and let \(M_{n}:=\left(\int_{a}^{b} f^{n}\right)^{1 / n}\). Show that \(\lim \left(M_{n}\right)=\sup \\{f(x): x \in[a, b]\\}\)

Short Answer

Expert verified
By proving by contradiction, we assume that the limit of the sequence \(M_{n}\) is less than the supremum of \(f(x)\) and we show that this assumption leads to a contradiction. Therefore, \( \lim M_{n} = \sup\{f(x) : x \in [a, b]\} \).

Step by step solution

01

Problem Setup

Given the hypotheses that \(f\) is a continuous, non-negative function on \([a, b]\) and the definition of \(M_{n}\), we want to show that \( \lim M_{n} = \sup\{f(x) : x \in [a, b]\} \). Formalize this by letting \(M = \sup\{f(x) : x \in [a, b]\}\). This ensures \(f(x) \leq M\) for all \(x \in [a, b]\).
02

Assumption Then Proof by Contradiction

Next, assume that for some \(\epsilon > 0, \lim M_{n} < M - \epsilon\). By continuity, there exists some \( x' \in [a, b]\) where \( f(x') > M - \epsilon\). This means \(\int_{a}^{b} f^n \geq (b - a) (M - \epsilon)^n\) and hence \((\int_{a}^{b} f^n)^{1/n} \geq (b - a)^{1/n} (M - \epsilon)\). Thus as \(n 鈫� 鈭瀄), we will have \((b - a)^{1/n} (M - \epsilon) > \lim M_{n}\), hence contradicting our assumption.
03

Conclusion

The contradiction established in step 2 tells us that our assumption that \(\lim M_{n} < M - \epsilon\) was wrong. As \(\epsilon\) was arbitrary, it means that \(\lim M_{n} \geq M\). But since we already know \(M_{n} \leq M\), we have \(\lim M_{n} = M\), completing the proof.

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Most popular questions from this chapter

If \(\alpha(x):=-x\) and \(\omega(x):=x\) and if \(\alpha(x) \leq f(x) \leq \omega(x)\) for all \(x \in[0,1]\), does it follow from the Squeeze Theorem \(7.2 .3\) that \(f \in \mathcal{R}[0,1]\) ?

Let \(B(x):=-\frac{1}{2} x^{2}\) for \(x<0\) and \(B(x):=\frac{1}{2} x^{2}\) for \(x \geq 0 .\) Show that \(\int_{a}^{b}|x| d x=B(b)-B(a)\).

Let \(f:[a, b] \rightarrow \mathbb{R}\). Show that \(f \notin \mathcal{R}[a, b]\) if and only if there exists \(\varepsilon_{0}>0\) such that for every \(n \in \mathbb{N}\) there exist tagged partitions \(\dot{\mathcal{P}}_{n}\) and \(\mathcal{Q}_{n}\) with \(\left\|\dot{\mathcal{P}}_{n}\right\|<1 / n\) and \(\left\|\dot{\mathcal{Q}}_{n}\right\|<1 / n\) such that \(\left|S\left(f: \dot{\mathcal{P}}_{n}\right)-S\left(f ; \dot{\mathcal{Q}}_{n}\right)\right| \geq \varepsilon_{0}\)

If \(f\) and \(g\) are continuous on \([a, b]\) and if \(\int_{a}^{b} f=\int_{a}^{b} g\), prove that there exists \(c \in[a, b]\) such that \(f(c)=g(c)\).

Suppose that \(f\) is bounded on \([a, b]\) and that there exists two sequences of tagged partitions of \(\\{a, b]\) such that \(\|\dot{\mathcal{P}}\| \rightarrow 0\) and \(\left\|\dot{\mathcal{Q}}_{n}\right\| \rightarrow 0\), but such that \(\lim _{n} S\left(f ; \dot{\mathcal{P}}_{n}\right) \neq \lim _{n} S\left(f ; \mathcal{Q}_{n}\right) .\) Show that \(f\) is not in \(\mathcal{R}[a, b]\).
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