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Chapter 10: Problem 5

If \(f_{k}(x):=k\) for \(x \in[1 / k, 2 / k]\) and \(f_{k}(x):=0\) elsewhere on \([0,2]\), show that \(f_{k}(x) \rightarrow 0\) but that \(\int_{0}^{2} f_{k}=1\)

Short Answer

Expert verified
The function sequence \(f_{k}(x)\) converges pointwise to 0, but the integral of each function over the interval [0,2] equals 1.

Step by step solution

01

Pointwise Convergence

For any selected value of \(x\), if we select \(k > 2/x\), then \(x \notin [1/k, 2/k]\), so \(f_{k}(x) = 0\). Since it can be done for any \(x\) and \(k\), then \(f_{k}(x) \rightarrow 0\) as \(k \rightarrow \infty\). In other words, we see that the function \(f_{k}(x)\) converges to 0 pointwise.
02

Integral Calculation

To find the integral of the function \(f_{k}\) in the interval [0,2], notice that as per the definition, \(f_{k}(x)=0\) for \(x\) outside the interval \([1/k,2/k]\), the integral can be evaluated only in this interval. Since in this interval \(f_{k}(x) = k\), the integral becomes \(\int_{1/k}^{2/k} k dx = k * (2/k - 1/k) = 1\). Hence the integral of \(f_{k}\) over [0,2] equals 1, for all \(k\).

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Most popular questions from this chapter

Show that the integral \(\int_{0}^{\infty} \sqrt{x} \cdot \sin \left(x^{2}\right) d x\) is convergent, even though the integrand is not bounded as \(x \rightarrow \infty\). [Hint: Make a substitution.]

If \(f_{n}(x):=x^{n}\) for \(n \in \mathbb{N}\), show that \(f_{n} \in \mathcal{L}[0,1]\) and that \(\left\|f_{n}\right\| \rightarrow 0\). Thus \(\left\|f_{n}-\theta\right\| \rightarrow 0\), where \(\theta\) denotes the function identically equal to \(0 .\)

Show that the following functions belong to \(\mathcal{R}^{*}[0,1]\) by finding a function \(F_{k}\) that is continuous on \([0,1]\) and such that \(F_{k}^{\prime}(x)=f_{k}(x)\) for \(x \in[0,1] \backslash E_{k}\), for some finite set \(E_{k}\). (a) \(f_{1}(x):=(x+1) / \sqrt{x} \quad\) for \(x \in(0.1]\) and \(f_{1}(0):=0\). (b) \(f_{2}(x):=x / \sqrt{1-x} \quad\) for \(x \in[0,1)\) and \(f_{2}(1):=0\). (c) \(f_{3}(x):=\sqrt{x} \ln x \quad\) for \(x \in(0,1]\) and \(f_{3}(0):=0\). (d) \(f_{4}(x):=(\ln x) / \sqrt{x} \quad\) for \(x \in(0,1]\) and \(f_{4}(0):=0\). (e) \(f_{5}(x):=\sqrt{(1+x) /(1-x)}\) for \(x \in[0,1)\) and \(f_{5}(1):=0\) (f) \(f_{6}(x):=1 /(\sqrt{x} \sqrt{2-x}) \quad\) for \(x \in(0,1]\) and \(f_{6}(0):=0\).

Apply Hake's Theorem to \(g(x):=(1-x)^{-1 / 2}\) for \(x \in[0,1)\) and \(g(1):=0\).

Let \(f \in \mathcal{R}^{*}[a, b]\), let \(g\) be monotone on \([a, b]\) and suppose that \(f \geq 0\). Then there exists \(\xi \in[a, b]\) such that \(\int_{a}^{b} f g=g(a) \int_{a}^{\xi} f+g(b) \int_{\xi}^{b} f .\) (This is a form of the Second Mean Value Theorem for integrals.)
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