Homework 9

Q1

As \(\lim_{x\to\infty} f(x) = a\), it follows that \(\int_0^x f(t)\ dt \to\infty\) as \(x\to\infty\). Then by l’Hopital’s rule and FTOC we have \[ \lim_{x\to\infty}\frac{\int_0^x f(t)\ dt}x = \lim_{x\to\infty} f(x) = a. \]

Q2

As \(g(x)\) does not change sign, for all \(x\in[a,b]\) \[ m\,g(x) \leq f(x)\,g(x) \leq M\,g(x) \] then \[ m \int_a^b g(x)\ dx \leq \int_a^b f(x)g(x)\ dx \leq M \int_a^b g(x)\ dx \] and the result follows.

Q3

For each \(n\in\mathbb{N}\) we have the following

\[\begin{align*} \int_0^\pi f(x)\left\lvert\sin nx\right\rvert dx &= \sum_{k=1}^n \int_{(k-1)\pi/n}^{k\pi/n} f(x)\left\lvert\sin nx\right\rvert dx \end{align*}\] and at each \(k\) by result of question 2, there is a \(\mu_k\in[a,b]\) such that \[ \mu_k\int_a^b \left\lvert\sin nx\right\rvert dx = \int_a^b f(x)\left\lvert\sin nx\right\rvert dx \] where \(a = \frac{(k-1)\pi}n, b = \frac{k\pi}n\). Additionally as \(f\) is continuous we know that \(\mu_k = f(c_k)\) for some \(c_k\in [a,b]\). Then \[\begin{align*} \int_0^\pi f(x)\left\lvert\sin nx\right\rvert dx &= \sum_{k=1}^n f(c_k)\int_{(k-1)\pi/n}^{k\pi/n} \left\lvert\sin nx\right\rvert dx \\ &= \sum_{k=1}^n f(c_k) \frac1n \int_{0}^{\pi} \left\lvert\sin x\right\rvert dx \\ &= \sum_{k=1}^n \frac2n f(c_k) \end{align*}\] then as we take the limit \[\begin{align*} \lim_{n\to\infty} \int_0^\pi f(x)\left\lvert\sin nx\right\rvert dx &= \frac2\pi \lim_{n\to\infty} \frac\pi{n} \sum_{k=1}^n f(x_k) \\ &= \frac2\pi \int_0^x f(x)\ dx. \end{align*}\]

Homework 10

Q1

Part (i). Integrating by parts, \[\begin{align*} I(m,n) &= \int_0^{\pi/2}\cos^m x\sin^n x\ dx \\ &= \left[ \frac{\cos^{m-1} x\sin^{n+1} x}{n+1} \right]^{\pi/2}_0 + \int_0^{\pi/2} \frac{m-1}{n+1} \cos^{m-2}x\sin^{n+2}x\ dx \\ &= \frac{m-1}{n+1} \int_0^{\pi/2} \cos^{m-2}x\sin^nx (1-\cos^2 x)\ dx \\ &= \frac{m-1}{n+1} (I(m-2, n) - I(m,n)) \end{align*}\] solving for \(I(m,n)\) gives \[ I(m,n) = \frac{m-1}{n+m} I(m-2,n) \] as desired. Proceed similarly for the case \(I(m, n-2)\), \[\begin{align*} I(m,n) &= \int_0^{\pi/2}\cos^m x\sin^n x\ dx \\ &= \left[ \frac{-\sin^{n-1} x\cos^{m+1} x}{m+1} \right]^{\pi/2}_0 + \int_0^{\pi/2} \frac{n-1}{m+1} \cos^{m+2}x\sin^{n-2}x\ dx \\ &= \frac{n-1}{m+1} (I(m, n-2) - I(m,n)) \end{align*}\] and solve for \(I(m,n)\) the same way.

Part (ii). To find \(I(2m,2n)\) repeatedly apply part (i). \[\begin{align*} I(2m,2n) &= I(0,2n) \prod_{i=1}^m \frac{2i-1}{2i+2n} \\ &= I(0,0) \prod_{i=1}^m \frac{2i-1}{2i+2n} \prod_{j=1}^n \frac{2j-1}{2j} \\ &= \frac\pi2 \frac{\prod_{i=1}^m (2i-1) \prod_{j=1}^n (2j-1)}{2^{m+n}(m+n)!} \end{align*}\] for the product terms observe that \[ \prod_{i=2}^m (2i-1) = \frac{(2m)!}{2^m m!} \] so \[ I(2m,2n) = \frac{(2m)!(2n)!}{2^{2(m+n)}(m+n)!} \frac\pi2 \]

Q2

Let \(f(x) = \frac{(-1)^{\lfloor x\rfloor}}{\sqrt{\lfloor x\rfloor}}\), then integrating \(f\) from \(a\) is (more or less) equivalent to adding the series \(\sum_{n=a}^\infty \frac{(-1)^n}{\sqrt{n}}\), which is convergent by alternating series test. However integrating \(f^2\) is equivalent to adding the harmonic series \(\frac1n\) which diverges.

Q3

Part (i). Let \(S_n(x)\) denote the partial sum \(\sum_{m=1}^n u_n(x)\). Let \(\mu = \inf_{x\in[a,b]} S(x)\) and for each \(n\in\mathbb{N}\), let \(X_n = \left\{x\in[a,b]: S_n(x) \leq \mu\right\}\) which is closed. Note that \(X_n\) is also nonempty as \(S_n\) achieves its minimum. Next observe that \(X_1\supseteq X_2\supseteq \dots\), since if \(x\in X_{n+1}\), then \(S_{n+1}(x) < \mu\), then \(S_n(x) + u_{n+1}(x) \leq \mu\) so \(x\in X_n\) too. By nested ball theorem \(\bigcap_{n=1}^\infty X_n\) is nonempty, let \(a\) be in the intersection. As each \(S_n(a) \leq \mu\), \(S(a) = \lim_n S_n(a) \leq \mu\) too which suffices.

Part (ii). \(S(x)\) could achieve its minimum on an endpoint, so the conclusion for \((a,b)\) fails. This can hold in a degenerate example where \(u_1(x) = x\) and \(u_n(x) = 0\) for all \(n>1\).

Homework 11

Q1

We show \(f_n(x)\) uniformly converges to \(f'(x)\) on any finite closed interval \([a,b]\), which will imply the result.

Consider the difference \[ f_n(x) - f'(x) = e^n\left[f(x+e^{-n}) - f(x)\right] - f'(x) \] by Taylor’s theorem there exists \(c\in (x,x+e^{-n})\) such that \[\begin{align*} f(x+e^{-n}) &= f(x) + f'(c)(e^{-n}) \\ f_n(x) - f'(x) &= f'(c) - f'(x) \end{align*}\] Now as \(f'\) is continuous, and thus uniformly continuous on \([a,b]\), by making \(n\) sufficiently large we can ensure that \(c\) and \(x\) get arbitrarily close. Specifically fix \(\varepsilon> 0\) and let \(\delta\) such that \(\left\lvert c-x\right\rvert<\delta\implies \left\lvert f'(c)-f'(x)\right\rvert<\varepsilon\), then choose \(N > -\ln\delta\) then whenever \(n\geq N\), \(e^{-n} < \delta\) and \(\left\lvert c-x\right\rvert<\delta\), then the result follows.

Q2

First we show that \(\sum_{n=1}^\infty u'_n(x_0)\) converges. By Cauchy criterion, we consider \(\left\lvert\sum_{n=k}^l u'_n(x_0)\right\rvert\). Note that for each \(n\in\left\{k,k+1,\dots,l\right\}\) there exists \(\delta_n\) such that \[ \left\lvert x-x_0\right\rvert<\delta_n \implies \left\lvert\frac{u_n(x) - u_n(x_0)}{x-x_0} - u'_n(x_0)\right\rvert < \frac{\varepsilon}{2(l-k+1)} \] then whenever \(\left\lvert x-x_0\right\rvert<\min_n \delta_n\), \[\begin{align*} \left\lvert\sum_{n=k}^l u'_n(x_0)\right\rvert &= \left\lvert\sum_{n=k}^l \frac{u_n(x) - u_n(x_0)}{x - x_0}\right\rvert + \frac\varepsilon 2 \end{align*}\] and uniform convergence of \(\sum_{n=1}^\infty \frac{u_n(x) - u_n(x_0)}{x - x_0}\) controls the other term.

Next we show \(f'(x_0) = \sum_{n=1}^\infty u'_n(x_0)\). Consider \[\begin{align*} \left\lvert \frac{\sum_n u_n(x) - \sum_n u_n(x_0)}{x-x_0} - \sum_n u'_n(x_0) \right\rvert &= \left\lvert \sum_n \frac{u_n(x) - u_n(x_0)}{x-x_0} - \sum_n u'_n(x_0) \right\rvert \end{align*}\] By property (2) for all \(x\) near \(x_0\), the tail of \(\sum_{n=N}^\infty \frac{u_n(x) - u_n(x_0)}{x-x_0}\) can be made arbitrarily small, and the same can be said for \(\sum_{n=N}^\infty u'_n(x_0)\) for some large enough \(N\), therefore \[\begin{align*} \left\lvert \frac{\sum_n u_n(x) - \sum_n u_n(x_0)}{x-x_0} - \sum_n u'_n(x_0) \right\rvert &\leq \left\lvert \sum_{n=1}^N \frac{u_n(x) - u_n(x_0)}{x-x_0} - \sum_{n=1}^N u'_n(x_0) \right\rvert + o(1) \\ &\leq \left\lvert \sum_{n=1}^N \frac{u_n(x) - u_n(x_0)}{x-x_0} - u'_n(x_0) \right\rvert + o(1) \end{align*}\] and the finite sum can be controlled using an argument similar to the earlier part.

Q3

We first show that \(\sum_{n=1}^\infty \int_a^\infty u_n(x)\ dx\) converges. By the Cauchy criterion consider \[\begin{align*} \left\lvert\sum_{n=k}^l \int_a^\infty u_n(x)\ dx\right\rvert &\leq\left\lvert \sum_{n=k}^l \int_a^b u_n(x)\ dx \right\rvert + \sum_{n=k}^l \left\lvert\int_b^\infty u_n(x)\ dx\right\rvert \end{align*}\] applying property (3) to \(g(b)\) controls the first term while applying property (1) controls the second summation.

Now it suffices to show that for any \(\varepsilon>0\), there exists large enough \(b\) such that \[ \left\lvert \int_a^b \sum_{n=1}^\infty u_n(x)\ dx - \sum_{n=1}^\infty \int_a^\infty u_n(x)\ dx \right\rvert< \varepsilon. \]

By property (2) we can exchange integral and sum \[ \int_a^b \sum_{n=1}^\infty u_n(x)\ dx = \sum_{n=1}^\infty \int_a^b u_n(x)\ dx \] Now \[\begin{align*} \sum_{n=1}^\infty\int_a^\infty u_n(x)\ dx &= \sum_{n=1}^K\int_a^\infty u_n(x)\ dx + o(1) \tag*{convergence} \\ &= \sum_{n=1}^K\int_a^b u_n(x)\ dx + o(1) \tag*{property (1)}\\ &= \sum_{n=1}^\infty\int_a^b u_n(x)\ dx + o(1) \tag*{property (3)} \end{align*}\] for sufficiently large \(K\) (chosen to satisfy property (3) in addition) and \(b\), which suffices.