(33.2) Let \(S\) be a nonempty bounded subset of \(\mathbb R\). For fixed \(c > 0\), let \(cS = \{ cs : s \in S \}\). Show that \(\sup(cS) = c\cdot\sup(S)\) and \(\inf(cS) = c\cdot\inf(S)\).
Proof. Let \(M = \sup S\). Then \(cM\) is an upper bound for \(cS\) as for all \(cs \in cS\) we have that \(M \ge s\) so \(cM \ge cs\) as \(c > 0\).
Now, if \(\alpha \ge cs\) for all \(cs \in cS\) (that is, \(\alpha\) is an upper bound for \(cS\)), then \(\alpha/c \ge s\) for all \(s \in S\) so \(\alpha\) is an upper bound for \(S\). But then \(M \le \alpha/c\) as \(M\) is the least upper bound for \(S\), so \(cM \le \alpha\). So \(cM = c\sup S = \sup(cS)\).
Now notice that \(\inf(cS) = -\sup(-cS) = -c\sup(-S) = c\inf S\). \(\Box\)
(33.6) Let \(f\) be integrable on \([a,b]\). Prove that, for any subset \(S \subseteq [a, b]\) we have
\[M(|f|, S) - m(|f|, S) \le M(f, S) - m(f, S)\]Hint. For \(x_0, y_0 \in S\), we have \(\vert f(x_0)\vert - \vert f(y_0)\vert \le \vert f(x_0) - f(y_0)\vert \le M(f, S) - m(f, S)\).
Proof. The first part of the hint follows from the triangle inequality, as
\[\vert f(x_0)\vert = \vert f(x_0) - f(y_0) + f(y_0) \vert \le \vert f(x_0) - f(y_0) \vert + \vert f(y_0) \vert.\]The second part of the hint comes from how \(M(f, S) \ge f(x_0)\) and \(m(f, S) \le f(y_0)\) for all \(x_0, y_0 \in S\).
On the other hand, we have that for any \(\epsilon > 0\), properties of the supremum and infimum guarantee that there exists \(x_0, y_0 \in S\) so that,
\[M(\vert f\vert, S) < \vert f(x_0)\vert + \epsilon/2 \quad\textrm{and} \quad m(\vert f\vert, S) > \vert f(y_0) \vert - \epsilon/2.\]So we have that,
\[M(\vert f\vert, S) - m(\vert f\vert, S) < \vert f(x_0)\vert - \vert f(y_0) \vert + \epsilon \le M(f, S) - m(f, S) + \epsilon.\]This holds for all \(\epsilon > 0\), so we conclude that the desired inequality holds. \(\Box\)
(33.5) Show that \(\left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert \le \frac{16\pi^3}{3}\).
Proof. Using Theorems 33.4 and 33.5, together with the FTC I (for example), we have that (as \(\left\vert \sin^8(e^x )\right\vert \le 1\))
\[\begin{align*} \left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \sin^8(e^x) \right\vert\; dx \\ &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \right\vert\; dx = \int_{-2\pi}^{2\pi} x^2 \; dx \\ &\le \frac{16\pi^3}3. \Box \end{align*}\](33.8) Let \(f\) and \(g\) be integrable functions on \([a, b]\).
It is a fact (see Exercise 33.7) that if \(h\) is integrable on \([a,b]\), then so is \(h^2\). Prove that \(fg\) is integrable on \([a, b]\). Hint. Use that \(4fg = (f+g)^2 - (f-g)^2\).
Show that \(\max(f,g)\) and \(\min(f,g)\) are integrable on \([a, b]\). You may use the results of Excercise 17.8 without proof.
Use that for integrable functions \(f, g\) we have that \(f+g\), \(f-g\), \((f+g)^2\), \((f-g)^2\), \(\vert f-g \vert\) are all integrable. Further applications of linearity of the integral prove (a), and together with \(\min(f,g) = \frac 12 (f+g) - \frac 12 \vert f-g \vert\) prove the first part of (b). The second part then follows from \(\max(f,g) = -\min(-f,-g)\) and integrability of \(-f\), etc.
(33.10) Let \(f\) be the function,
\[f(x) = \left\{\begin{array}{lr}\sin(\frac 1x) & x \ne 0\\0 & x=0\end{array}\right.\]Prove that \(f\) is integrable on \([-1, 1]\). Hint. See Excercise 33.11(c) and its solution in the textbook. (Why can we not apply the Dominated Convergence Theorem to prove that \(f\) is integrable?)
Proof. Let \(\epsilon > 0\). Notice that \(f\) is continuous on each of \([-1, -\epsilon/8]\) and \([\epsilon/8, 1]\), and is hence integrable. This means that there exist partitions \(Q_1\) of \([-1, -\epsilon/8]\) and \(Q_2\) of \([\epsilon/8, 1]\) so that
\[U(f, Q_1) - L(f, Q_1) < \epsilon/4 \quad\text{and}\quad U(f, Q_2) - L(f, Q_2) < \epsilon/4.\]Notice that \(P = Q_1 \cup Q_2\) is a partition of \([-1, 1]\) and furthermore that
\[U(f, P) = U(f, Q_1) + \epsilon/4 + U(f, Q_2)\]and,
\[L(f, P) = L(f, Q_1) - \epsilon/4 + L(f, Q_2).\](Why? The only terms in the two sums are those contributed by the interval \([-\epsilon/8, \epsilon/8]\) but we know that the supremum and infimum of \(f\) on any such interval are \(1\) and \(-1\) respectively and that the width of this interval is \(\epsilon/4\)) So,
\[\begin{align*} U(f, P) - L(f, P) &= (U(f, Q_1) + \epsilon/4 + U(f, Q_2)) - (L(f, Q_1) - \epsilon/4 + L(f, Q_2)) \\ &= (U(f, Q_1) - L(f, Q_1)) + (U(f, Q_2) - L(f, Q_2)) + \epsilon/2 \\ &< \epsilon/4 + \epsilon/4 + \epsilon/2 = \epsilon. \end{align*}\]We can find any such \(P\) for any \(\epsilon > 0\), so we conclude that \(f\) is integrable on \([-1, 1]\).