Supremum equation proof.

Question

I want to show that $$\overbrace{\sup\{f(x)-f(y):x,y\in X\}}^{(1)}=\overbrace{\sup\{|f(x)-f(y)|:x,y\in X\}}^{(2)}$$ where $f:X\to\mathbb{R}$ is a bounded function.

It seems obvious, but I don't quite know how to prove it.

Here is what I have so far:
I need to show that $(1)\leq (2)$ and $(1)\geq (2)$, which implies that $(1)=(2)$.
$(1)\leq (2):$
Let's take some $x_0, y_0\in X$, then $$ f(x_0)-f(y_0)\leq |f(x_0)-f(y_0)|\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$ $$ f(x_0)-f(y_0)\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$

Now I would think that

$$\sup\{f(x)-f(y):x,y\in X\}\leq \sup\{|f(x)-f(y)|:x,y\in X\}$$

Is that correct?

$(2)\leq (1):$

I don't have ideas here.

How to prove it? Any ideas or tips?

Answer 1

$\boxed{\leq \text{inequality}}$ Note that $$ f(x)-f(y)\leq\left|f(x)-f(y)\right| $$ holds for all pairs $(x,y)$. Therefore, taking the supremum of both sides, $$ \sup_{x,y} \left\{ f(x)-f(y) \right\} \leq \sup_{x,y} \left| f(x)-f(y) \right| $$

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$\boxed{\geq \text{inequality}}$ Since $f$ is bounded, we know that for each $\epsilon > 0$, we can find $(u,w)$ such that $$ \sup_{x,y} \left| f(x)-f(y) \right| \leq \left| f(u)-f(w) \right| + \epsilon. $$ Now, define the pair $(u^\prime,w^\prime)$ by $$ (u^{\prime},w^{\prime})=\begin{cases} (u,w) & \text{if }f(u)\geq f(w)\\ (w,u) & \text{otherwise} \end{cases} $$ so that $$ \left| f(u)-f(w) \right| = f(u^\prime) - f(w^\prime). $$ What can you conclude? (remember, $\epsilon$ is arbitrary)

Answer 2

Note that for every $x,y \in X$, we have (1) $\geq f(x)-f(y)$. Therefore, given $a,b\in X$, (1) $\geq f(a)-f(b)$ and (1) $\geq f(b)-f(a) =-(f(a)-f(b))$. This implies (1) $\geq |f(a)-f(b)|$. Now, we conclude (1) is an upper bound of $\{|f(x)-f(y)|:x,y\in X\}$.

Since (2) is the supremum, we have the inequality desired.

Answer 3

There's a mildly distracting complication: Let $Y = \{f(x) : x\in X\}$ be the image of the function $X.$

Thus instead of calling our set $\{f(x)-f(y) : x,y\in X\},$ we can call it $\{a-b : a,b\in Y\}.$ In other words, get rid of the function $f$ and state the proposition like this: $$ \sup\{|a-b| : a,b\in Y\} = \sup\{a-b: a,b\in Y\}. $$

We have $$ \{|a-b|:a,b\in Y\} = \{a-b: a,b\in Y\ \&\ a\ge b\} \subseteq \{a-b: a,b\in Y\}. \tag {inclusion} $$ Do you know how to prove that this inclusion implies that $$ \sup\{|a-b|:a,b\in Y\} \le \sup\{a-b:a,b\in Y\} \text{ ?} $$

The other inequality follows from $a-b\le |a-b|.$