1

I would like to show that for a metric space $(X, d)$ and a function $f \colon X \to \mathbb R$, $\sup\,\{f(x) \mid x \in X\}-\inf\,\{f(x) \mid x \in X\}=\sup\,\{|f(x)-f(y)| \mid x, y \in X\}$.

So far I have:

Since for all $x, y \in X$, $f(x)-f(y)\leq|f(x)-f(y)|$, we have that

\begin{align*} \sup\,\{|f(x)-f(y)| \mid x, y \in X\}&\geq \sup\,\{f(x)-f(y) \mid x, y \in X\} \\ &=\sup\,\{f(x) \mid x \in X\}+\sup\,\{-f(y) \mid y \in X\} \\ &=\sup\,\{f(x) \mid x \in M\}-\inf\,\{f(y) \mid y \in X\} \\ &=\sup\,\{f(x) \mid x \in X\}-\inf\,\{f(x) \mid x \in X\}. \end{align*}

I now want to show that

$\sup\,\{f(x) \mid x \in X\}-\inf\,\{f(x) \mid x \in X\} \geq \sup\,\{|f(x)-f(y)| \mid x, y \in X\}$.

but am in need of a little assistance. Any help is really appreciated.

user83343
  • 45

1 Answers1

1

Suppose $x_0,y_0\in X$.

If $f(x_0)\geq f(y_0)$, then $\lvert f(x_0)-f(y_0)\rvert=f(x_0)-f(y_0)$. But $$ f(x_0)\leq\sup\{f(x)\mid x\in X\}\qquad\text{and}\qquad f(y_0)\geq\inf\{f(y)\mid y\in X\}, $$ and so $$ \lvert f(x_0)-f(y_0)\rvert=f(x_0)-f(y_0)\leq\sup\{f(x)\mid x\in X\}-\inf\{f(y)\mid y\in X\}. $$ If $f(x_0)<f(y_0)$, then reversing their roles leads to the same inequality.

So, we have that $\sup\{f(x)\mid x\in X\}-\inf\{f(y)\mid y\in X\}$ is an upper bound on $\{\lvert f(x)-f(y)\rvert\mid x,y\in X\}$, and therefore it must be at least as large as the least upper bound.

Nick Peterson
  • 32,430