Upper and lower integration inequality

Question

I would like to learn how to prove that the following inequality holds.

Let $F$ be a bounded function on an interval $[a,b]$, so that there exists $B\geq 0$ such that $|f(x)| \leq B$ for every $x\in [a,b]$.

Show that $[ U(f^2,P) -L(f^2,P) \leq 2B [ U(f,P) -L(f,P) ] ]$ for all partitions $P$ of $[a,b]$.

Answer 1

Notation. For any finction $\varphi:[a,b]\to\mathbb{R}$ we denote $$ M_{[\alpha,\beta]}(\varphi)=\sup\limits_{x\in[\alpha,\beta]}\varphi(x)\qquad m_{[\alpha,\beta]}(\varphi)=\inf\limits_{x\in[\alpha,\beta]}\varphi(x) $$ where $[\alpha,\beta]\subset[a,b]$. For a given partition $P=\{\Delta_1,\ldots,\Delta_n\}$ of $[a,b]$ we denote $$ U(f,P)=\sum\limits_{i=1}^n M_{\Delta_i}(f)|\Delta_i|\qquad L(f,P)=\sum\limits_{i=1}^n m_{\Delta_i}(f)|\Delta_i| $$

Proposition. Let $f:[a,b]\to\mathbb{R}$ be a real valued bounded function such that $|f(x)|\leq B$ for all $x\in[a,b]$. Let $P$ be partition of $[a,b]$, then $$ U(f^2,P)-L(f^2,P)\leq 2B(U(f,P)-L(f,P)) $$

Proof For each $i\in\{1,\ldots,n\}$ we have $$ \begin{align} M_{\Delta_i}(f^2)-m_{\Delta_i}(f^2) &=M_{\Delta_i}(|f|^2)-m_{\Delta_i}(|f|^2)\\ &=M_{\Delta_i}(|f|)^2-m_{\Delta_i}(|f|)^2\\ &=(M_{\Delta_i}(|f|)+m_{\Delta_i}(|f|))(M_{\Delta_i}(|f|)-m_{\Delta_i}(|f|))\\ &\leq 2B(M_{\Delta_i}(|f|)-m_{\Delta_i}(|f|))\\ &\leq 2B(M_{\Delta_i}(f)-m_{\Delta_i}(f))\\ \end{align} $$ Hence $$ \begin{align} U(f^2,P)-L(f^2,P) &=\sum\limits_{i=1}^n(M_{\Delta_i}(f^2)-m_{\Delta_i}(f^2))|\Delta_i|\\ &\leq\sum\limits_{i=1}^n 2B(M_{\Delta_i}(f)-m_{\Delta_i}(f))|\Delta_i|\\ &2B(U(f,P)-L(f,P)) \end{align} $$

Remark. This proof used several standard properties of $M_\Delta$ and $m_\Delta$ functions. These properties are proved below.

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Lemma. Let $\varphi:[a,b]\to\mathbb{R}$ be a real valued bounded function such that $|\varphi(x)|\leq K$ for all $x\in[a,b]$. Let $\Delta\subset[a,b]$ then

1. $M_\Delta(\varphi)-m_\Delta(\varphi)=\sup_{x,y\in\Delta}|\varphi(x)-\varphi(y)|\tag{1}$
2. $M_\Delta(|\varphi|)-m_\Delta(|\varphi|)\leq M_\Delta(\varphi)-m_\Delta(\varphi)\tag{2}$
3. $M_\Delta(|\varphi|)\leq K\qquad m_\Delta(|\varphi|)\leq K\tag{3}$

Proof. 1) For all $x,y\in\Delta$ we have $m_\Delta(\varphi)\leq \varphi(x)\leq M_\Delta(\varphi)$ and $m_\Delta(\varphi)\leq \varphi(y)\leq M_\Delta(\varphi)$, so $$ |f(x)-f(y)|\leq M_\Delta(\varphi)-m_\Delta(\varphi)\tag{*} $$ Fix $\varepsilon>0$, then there exist $\tilde{x}\in\Delta$, $\tilde{y}\in\Delta$ such that $M_\Delta(\varphi)<\varphi(\tilde{x})+\varepsilon/2$ and $m_\Delta(\varphi)>\varphi(\tilde{y})-\varepsilon/2$. Hence $$ M_\Delta(\varphi)-m_\Delta(\varphi)<\varphi(\tilde{x})-\varphi(\tilde{y})+\varepsilon\tag{**} $$ Since $\varepsilon>0$ is arbtrary $(\,^{*})$ and $(\,^{**})$ implies $$ M_\Delta(\varphi)-m_\Delta(\varphi)=\sup\limits_{x,y\in\Delta}|\varphi(x)-\varphi(y)| $$

2) For all $x,y\in\Delta$ we have $||\varphi(x)|-|\varphi(y)||\leq|\varphi(x)-\varphi(y)|$, hence using $(1)$ we get $$ M_\Delta(|\varphi|)-m_\Delta(|\varphi|)=\sup\limits_{x,y\in\Delta}||\varphi(x)|-|\varphi(y)||\leq\sup\limits_{x,y\in\Delta}|\varphi(x)-\varphi(y)|=M_\Delta(\varphi)-m_\Delta(\varphi) $$

3) Since $|\varphi(x)|\leq K$ for all $x\in\Delta$, then $$ m_\Delta(|\varphi|)\leq M_\Delta(|\varphi|)\leq K $$

Lemma. Let $\varphi:[a,b]\to[c,d]$ be a real valued function. Let $\psi:[c,d]\to\mathbb{R}$ be continuous non decreasing function. Then for any interval $\Delta\subset[a,b]$ we have
$M_\Delta(\psi\circ \varphi)=\psi(M_\Delta(\varphi))\qquad m_\Delta(\psi\circ\varphi)=\psi(m_\Delta(\varphi))\tag{4}$

Proof. For all $x\in\Delta$ we have $\varphi(x)\leq M_\Delta(\varphi)$. Since $\psi$ is a non decreasing function, then for all $x\in\Delta$ we get $\psi(\varphi(x))\leq\psi(M_\Delta(\varphi))$. This implies $$ M_\Delta(\psi\circ\varphi)\leq \psi(M_\Delta(\varphi))\tag{***} $$ Fix $\varepsilon>0$. Since $\psi$ is continuous and non decreasing there exist $\delta>0$ such that $\psi(M_\Delta(\varphi))-\psi(t)<\varepsilon$ for all $M_\Delta(\varphi)-\delta<t<M_\Delta(\varphi)$. Moreover from definition of supremum we can find $\tilde{x}\in\Delta$ such that $M_\Delta(\varphi)-\delta<\varphi(\tilde{x)}<M_\Delta(\varphi)$. Now we set $t=\varphi(x)$ to get $$ \psi(M_\Delta(\varphi))<\psi(\varphi(\tilde{x}))+\varepsilon\tag{****} $$ Since $\varepsilon>0$ is arbitrary from $(\,^{***})$ and $(\,^{****})$ we get $$ M_\Delta(\psi\circ \varphi)=\psi(M_\Delta(\varphi)) $$ The proof of the other equality is similar.