Week 4 Worksheet Solutions

Home | Assessment | Notes | Worksheets | Tutorials

1. It suffices to prove that that $$|1_B-1_C|=1_{B\mathrm{△}C}$$ because of the way in which the integral of a simple function is defined. But $$|1_B-1_C|(x)=\{\begin{array}{ll}0& x\in B\cap C\\ 1& x\in B\setminus C\\ 1& x\in C\setminus B\\ 0& x\in X\setminus B\cup C\end{array}$$ is precisely the indicator function of $B\mathrm{△}C$.
2. In this context every space ${\mathcal{L}}^{\mathsf{2}}$ is just ${\mathbb{R}}^{\{1,2\}}$ and $$\int f\mathsf{d}\mu =f(1)+f(2)$$ so Minkowski's inequality says $$\begin{array}{rl}& {(|f(1)+g(1){|}^p+|f(2)+g(2){|}^p)}^{1/p}\\ \le & {(|f(1){|}^p+|f(2){|}^p)}^{1/p}+{(|g(1){|}^p+|g(2){|}^p)}^{1/p}\end{array}$$ and Holder's inequality says $$\begin{array}{rl}& |f(1)g(1)|+|f(2)g(2)|\\ \le & (|f(1){|}^p+|f(2){|}^p{)}^{1/p}(|g(1){|}^q+|g(2){|}^q{)}^{1/q}\end{array}$$
3. Let $\mu$ be counting measure on $(X,\mathcal{P}(X))$. The Lebesgue space ${\mathsf{L}}^{\mathsf{p}}(X,\mathcal{P}(X),\mu )$ is often denoted simply ${\ell }^{\mathsf{p}}(X)$.
   1. As $\{1,\dots ,N\}$ is finite, integration with respect to counting measure is in this case a finite sum, and $f$ belongs to ${\mathcal{L}}^{\mathsf{p}}$ in this case if and only if $f$ does not take the value $\mathrm{\infty }$ regarless of what $p$ is.
   2. Put $f(n)=n^{-\alpha }$. For $f$ to belong to ${\ell }^q$ we need $$\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^{q\alpha }}}<\mathrm{\infty }$$ which happens if $q\alpha >1$. For $f$ to fail to belong to ${\ell }^p$ we need $$\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^{p\alpha }}}=\mathrm{\infty }$$ which will happen if $p\alpha <1$. We will satisfy both requirements if e.g. $\alpha =2/(p+q)$.
   3. Fix $f\in {\ell }^p$. This means $$\sum _{n=1}^{\mathrm{\infty }}|f(n){|}^p$$ is finite and in particular there is $N\in \mathbb{N}$ such that $|f(n)|<1$ whenever $n\ge N$. For such $n$ we have $$|f(n){|}^q\le |f(n){|}^p$$ and therefore $$\sum _{n=1}^{\mathrm{\infty }}|f(n){|}^q$$ automatically converges to a finite value.
4. If $p=q$ there is nothing to prove so assume $p>q$. We apply Holder's inequality with $$P={\displaystyle \frac{p}{q}}Q={\displaystyle \frac{p}{p-q}}$$ to get $$\int |f{|}^q\cdot 1\mathsf{d}\mu \le \Vert |f{|}^q{\Vert }_P\Vert 1{\Vert }_Q=\Vert f{\Vert }_p^q\cdot \mu (X{)}^{1/Q}<\mathrm{\infty }$$ whence $f$ belongs to ${\mathsf{L}}^{\mathsf{q}}(X,\mathcal{B},\mu )$.
5. In one case the measure is a counting measure and in the other it is finite. For finite measures the constant functions belongs to all Lebesgue spaces and may be used in applications of Hölder's inequality. For counting measures being in a Lebesgue space forces decay at infinity which was crucial in Problem 3.
6. 1. Suppose that $b$ belongs to ${\ell }^{\mathsf{p}}(\mathbb{N})$ which means $$\sum _{n=1}^{\mathrm{\infty }}|b(n){|}^p<\mathrm{\infty }$$ and note then that $$\sum _{n=1}^{\mathrm{\infty }}|(Tb)(n){|}^p=\sum _{n=2}^{\mathrm{\infty }}|b(n){|}^p\le \sum _{n=1}^{\mathrm{\infty }}|b(n){|}^p<\mathrm{\infty }$$ so $T(b)$ also belongs to ${\ell }^{\mathsf{p}}$.
   2. Take $b(n)=2^{-n}$. We have $$\Vert b{\Vert }_{\mathsf{1}}=1\Vert T(b){\Vert }_{\mathsf{1}}={\displaystyle \frac{1}{2}}$$ so $T$ does not preserve the norms of elements in e.g. ${\ell }^{\mathsf{1}}(\mathbb{N})$ and is therefore not an isometry.
7. Define $\rho :\mathsf{C}(I_n)\to \mathbb{R}$ by $$\rho (f)=\underset{n}{\overset{n+1}{\int }}f(x)\mathsf{d}x$$ and note, by properties of the Riemann integral, that $\rho$ is linear and satisfies $$|\rho (f)|\le \Vert f{\Vert }_{\mathsf{u}}$$ for all $f\in \mathsf{C}(I_n)$. Moreover, when $f\ge 0$ we have $\rho (f)\ge 0$. There is therefore - by the Riesz representation thorem - a measure ${\mu }_n$ on the Borel subsets of $I_n$ such that $$\underset{n}{\overset{n+1}{\int }}f(x)\mathsf{d}x=\int f\mathsf{d}{\mu }_n$$ for all $f\in \mathsf{C}(I_n)$.
   
   By properties of the Riemann integral, we have for every compactly supported continuous function $f:\mathbb{R}\to \mathbb{R}$ that $$\int f\mathsf{d}\lambda =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\mathsf{d}x=\sum _{n\in \mathbb{Z}}\int f\mathsf{d}{\mu }_n$$ with the indefinite Riemann integral existing because $f=0$ outside some large interval $[-A,A]$. By suitably approximating $1_{(a,b)}$ by continuous functions for all $a<b\in \mathbb{R}$ one can show that $$\lambda ((a,b))=\sum _{n\in \mathbb{Z}}{\mu }_n((a,b))$$ for all such intervals.