A property of "function amplitude."

Question

Let $P=[a_1,b_1]\times\cdots\times[a_n,b_n]$ be a rectangle, $x\in P.$

Definition Let $f:P\to \mathbb{R}$ be a bounded function,then define the amplitude of function $f$ at the point $x$ as $$\omega_f(x):=\lim_{r\to 0+}\sup\left\{|f(x_1)-f(x_2)|:\ x_1,x_2\in B(x,r)\cap P\right\}.$$ Here $B(x,r)$ represents the open ball centered at point $x$ with a radius of $r.$

If we define the function $$B:P\to \mathbb{R},\ x\mapsto \lim_{r\to 0+}\sup_{a\in B(x,r)\cap P} f(a),$$ and function $$A:P\to \mathbb{R},\ x\mapsto \lim_{r\to 0+}\inf_{a\in B(x,r)\cap P} f(a),$$ then we can see $$\omega_f(x)=B(x)-A(x),\ \forall\ x\in P.$$ So each interval $[A(x),B(x)]$ corresponds to the amplitude of function $f$ at the point $x$.

Define the Supremum and infimum of function $f$ as $$m:=\inf_{x\in P} f(x),\ M:=\sup_{x\in P} f(x),$$ then I can guess that there is $$[m,M]\subset\bigcup_{x\in P} [A(x),B(x)].$$

In fact, I'm planning to use this conclusion to prove a theorem about Riemann integral($f$ is Riemann-integrable iff the region under its graph is Jordan-measurable),but I meet great difficulties in proving this conclusion, since I don't have enough analytical techniques.

Question: Prove that $$[m,M]\subset\bigcup_{x\in P} [A(x),B(x)].$$ Any hint is also welcomed.

Answer 1

I. Show $[m,M] \supset \cup_x [A(x),B(x)]$

Take any $y\in \cup_x [A(x),B(x)]$. Then it must be in $[A(x),B(x)]$ for some $x$.

Hence, for any $y\in\cup_x [A(x),B(x)]$, there is an $x$ such that: $$\inf_x f(x)\leq \lim_{r\rightarrow0}\inf_{z\in B_r(x)} f(z)\leq y \leq \lim_{r\rightarrow0}\sup_{z\in B_r(x)} f(z) \leq \sup_x f(x).$$

Thus $y\in[m,M]$.

So we have that $[m,M] \supset \cup_x [A(x),B(x)].$

II. Show $[m,M] \subset \cup_x [A(x),B(x)]$

Assume that $y\in[m,M]$ and the contrary that $\forall x$ either $y<A(x)$ or $B(x)<y$. So we can partition $P$ into two subsets $X_a$ and $X_b$ where $y<A(x)$ for $x\in X_a$ and $B(x)<y$ for $x\in X_b$ with $P=X_a\cup X_b$.

We have that $m=\min\{ \inf_{X_a}f, \inf_{X_b} f \}$ and $M=\max\{ \sup_{X_a}f, \sup_{X_b} f \}$.

Assuming that neither $X_a$ nor $X_b$ is empty, we get that $$m=\inf_{X_b}f\leq\sup_{X_b}f<y<\inf_{X_a}f \leq\sup_{X_a}f =M$$

But $P$ is connected so we can find $z\in\overline{X}_b\cap X_b$ (without loss of generality) such that $B_r(z)\cap X_a \neq \emptyset$ and $B_r(z)\cap X_b \neq \emptyset$ for all $r$. I.e. one of the partition sets contains a point on its "inner boundary" with the other partition set. This implies that $A(z)\leq \sup_{X_b} f$ and that $\inf_{X_a} f \leq B(z)$.

So we conclude that $y\in[A(z), B(z)]$ which is a contradiction.

If either $X_a$ or $X_b$ is empty, then the result changes slightly. I'll assume $X_a=\emptyset$ without loss of generality.

$$M=\sup_{X_b}f<y$$

Thus $y\notin [m,M]$, a contradiction.

The conclusion is that $[m,M] = \cup_x [A(x),B(x)]$.

Answer 2

Inspired by @jdods, finally I obtain the following theorem:

Theorem: Let $(P,\mathcal{T})$ be a connected topology space, and $f:P\to \mathbb{R}$ be a bounded function, define $$m:=\inf_{x\in P} f(x),\ M:=\sup_{x\in P} f(x),$$ and define functions $$A:P\to \mathbb{R},\ x\mapsto \sup_{\text{$U$ is a open neighborhood of $x$}} \big[\inf_{a\in U} f(a)\big],$$ $$B:P\to \mathbb{R},\ x\mapsto \inf_{\text{$U$ is a open neighborhood of $x$}} \big[\sup_{a\in U} f(a)\big],$$ then we have $$(m,M)\subset\bigcup_{x\in P} [A(x),B(x)]\subset [m,M].\ (*)$$ And if topology space $(P,\mathcal{T})$ is also compact, then we have $$\bigcup_{x\in P} [A(x),B(x)]=[m,M].\ (**)$$

Proof of $(*)$: It's obvious that $\bigcup_{x\in P} [A(x),B(x)]\subset [m,M].$ Let's prove that $(m,M)\subset \bigcup_{x\in P} [A(x),B(x)].$

For arbitrary $y\in (m,M),$ if we suppose that $y\notin \displaystyle\bigcup_{x\in D}[A(x),B(x)],$ then we have $$y\in (-\infty,A(x)]\cup[B(x),+\infty),\ \forall\ x\in X.$$

Define sets $$X_a:=\left\{x\in X| y< A(x)\right\},\ X_b:=\left\{x\in X| y> B(x)\right\},$$ then $P=X_a\cup X_b.$ Let's prove that $X_a,X_b\neq \emptyset.$

In fact, if $X_a=\emptyset,$ then $P=X_{b},$ thus $y\geqslant B(x),\ \forall\ x\in P.$

But $$B(x)\geqslant f(x),\ \forall\ x\in X,$$ thus $y\geqslant f(x),\ \forall\ x\in X,$ so $y\geqslant M,$ which is a contradiction. Now we see that $X_a\neq \emptyset,$ and similarly we can prove $X_b\neq \emptyset.$

Since $X_a\cup X_b=X,$ and $X_a, X_b\neq \emptyset,$ thus we have $\overline{X_a},\overline{X_b}\neq \emptyset,$ and $\overline{X_a}\cup \overline{X_b}=X.$ From the connectness of $P$ we infer that $\overline{X_a}\cap \overline{X_b}\neq \emptyset.$

Take $x_0\in \overline{X_a}\cap \overline{X_b},$ and assume that $U$ is an open neighborhood of $x_0,$ then we have $(U\cap \overline{X_a})\neq \emptyset,\ (U\cap \overline{X_b})\neq \emptyset.$ Take $x_1\in (U\cap \overline{X_a}),\ x_2\in (U\cap \overline{X_b}),$ then $U$ is an open neighborhood of both $x_1$ and $x_2.$ Thus $$\inf_{a\in U} f(a)\leqslant A(x_2),\ B(x_1)\leqslant \sup_{a\in U} f(a).$$ Notice that $$A(x_2)\leqslant B(x_2)< y<A(x_1)\leqslant B(x_1),$$

we have $$\inf_{a\in U} f(a)<y<\sup_{a\in U} f(a). $$ From the arbitrariness of $U$ we infer that $$A(x_0)\leqslant y\leqslant B(x_0),$$ which is a contradiction.

So now we've proved that $$(m,M)\subset\bigcup_{x\in P} [A(x),B(x)]\subset [m,M].\ (*)$$

To prove $(**)$,the second part of the theorem, we need to prove some useful propositions first:

Definition 1：Let $X$ be a topology space,，$x_0\in X,$ $f:X\to \mathbb{R}$ be a real-valued function. If $\forall\ \varepsilon>0,$ $\ \exists $ an open neighborhood of $x_0$,denoted as $U,$ such that $$f(x)<f(x_0)+\varepsilon,\ \forall\ x\in U,$$ then we say function $f:X\to \mathbb{R}$ is upper semi-continuous at point $x_0$. If function $f$ is upper semi-continuous at every point in $X$ , then we say $f$ is upper semi-continuous function.

Similarly, If $\forall\ \varepsilon>0,$ $\ \exists $ an open neighborhood of $x_0$,denoted as $U,$ such that $$f(x)>f(x_0)-\varepsilon,\ \forall\ x\in U,$$ then we say function $f:X\to \mathbb{R}$ is lower semi-continuous at point $x_0$. If function $f$ is lower semi-continuous at every point in $X$ , then we say $f$ is lower semi-continuous function.

We can see that if both $f$ and $g$ are upper semi-continuous, then function $f+g,$ is also upper semi-continuous, function $-f$ is lower semi-continuous. And if upper semi-continuousfunction $f$ is always negative on $X$ or always positive on $X,$ then function $1\over{f}$ is lower semi-continuous.

Now we define a set as $$T:=\left\{(-\infty,a)|\ a\in \mathbb{R}\right\}\cup \left\{\mathbb{R}\right\}\cup \left\{\emptyset\right\},$$ then we can see $T$ is a topology on $\mathbb{R},$ and the following definition is equivalent to definition 1:

Definition 2: Let $X$ be a topology space, $x_0\in X,$ $f:X\to \mathbb{R}$ be a real-valued function. If map $f:X\to (\mathbb{R}，T)$ is continuous at point $x_0,$ then we say function $f:X\to \mathbb{R}$ is upper semi-continuous at point $x_0$. If map $f:X\to (\mathbb{R}，T)$ is continuous , then we say function $f:X\to \mathbb{R}$ is upper semi-continuous.

The following lemma is important:

Lemma: If topological space $(X,\mathcal{T})$ is compact, and function $f:X\to \mathbb{R}$ is upper semi-continuous, then $f$ is bounded above, and $f$ has a maximum on $X.$

Proof: Define set $$X_n:=\left\{x\in X|\ f(x)<n\right\},$$ here $n\in \mathbb{N},$ then $X=\displaystyle\bigcup_{n\in \mathbb{N}} X_n,$ and according to Definition 2, $X_{n}$ is open set in $X.$ From the compactness of $X$ we can see there exist $n_1,\cdots,n_k,$ such that $$X=\bigcup_{i=1}^k X_{n_i}.$$ Let $M'=\displaystyle\max\left\{n_1,\cdots,n_k\right\},$ then $X=X_{M'},$ thus $f(x)<M',\ \forall\ x\in X,$ so $f$ is bounded above.

Now let's prove that if upper semi-continuous function $f$ is bounded above, then $f$ has a maximum $X$.

In fact, suppose that $f$ has no maximum, then define $K:=\sup_{x\in X} f(x),$ we can see function $g:={1\over{K-f}}:X\to \mathbb{R}$ is upper semi-continuous, thus $g$ is bounded above. Let $M$ be a upper bound of $g$, then $$g(x)={1\over{K-f(x)}}\leqslant M,\ \forall\ x\in X.$$ Notice that $g(x)>0,\ \forall\ x\in X,$ so $M>0,$ thus $$K-f(x)\geqslant {1\over{M}},\ \forall\ x\in X,$$ i.e. $$f(x)\leqslant K-{1\over{M}},\ \forall\ x\in X,$$ which is a contradiction. Q.E.D

Similarly,If topological space $(X,\mathcal{T})$ is compact, and function $f:X\to \mathbb{R}$ is lower semi-continuous, then $f$ is bounded below, and $f$ has a minimum on $X.$

Now let's complete the proof of the theorem:

Proof of $(**)$: First we'll prove that function $B(x)$ is upper semi-continuous, and function $A(x)$ is lower semi-continuous.

In fact, according to the definition of $B(x),$ for any $\varepsilon>0,$ there exists an open neighborhood of $x,$ such that $$\sup_{a\in U} f(a)<B(x)+\varepsilon.$$ Then assume that $t\in U,$ then $U$ is also an open neighborhood of $t,$ so $$B(t)\leqslant \sup_{a\in U} f(a).$$ Now we see that $$B(t)<B(x)+\varepsilon,\ \forall\ t\in U.$$ Thus function $B(x)$ is upper semi-continuous. Similarly we can prove that function $A(x)$ is lower semi-continuous.

Now notice that $$\sup_{x\in P} B(x)=M, \ \inf_{x\in P} A(x)=m,$$ and since $P$ is compact, according to the lemma we have $$m\in \bigcup_{x\in P} \left\{A(x)\right\},\ M\in \bigcup_{x\in P} \left\{B(x)\right\}.$$ Q.E.D