Suppose $f$ is bounded on $(a,b)$.
Is $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}$=$\sup\{f(x):a\leq x\leq b\}-\inf\{f(x):a\leq x\leq b\}=M-m$ (M being the sup and m the inf)
Possible proof:
For all $|f(x')-f(x'')|$ we know that $m\leq f(x'),f(x'')\leq M$. Hence, $-(M-m) \leq f(x')-f(x'')\leq M-m$ which implies $|f(x')-f(x'')|\leq M-m$. Therefore, $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}\leq M-m$.
Now I'm stuck as to how to show the other inequality. Is the statement even true (counter-examples?) or can anyone show me how to proceed with the proof.
Edit:
Suppose $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}< M-m$. Hence, $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}+m < M$
Therefore, there exists $f(x'), a\leq x'\leq b$ s.t $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}+m< f(x')$. Also, $m< f(x')-\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}$
Similarly, there exists $f(x''), a\leq x''\leq b$ s.t $f(x'') < f(x')-\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}$
This implies that $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}<f(x')-f(x'')$ which is a contradiction.
Hence, $\sup\{|f(x')-f(x'')|:a\leq x',x''\leq b\}=M-m$
Is this proof correct?
For $a\leq x\leq b$ and $a\leq y\leq b$, we have $f(x)-f(y)\leq M-f(y)\leq M-m$, and we also have $f(y)-f(x)\leq M-m$ with the same reason, so $|f(x)-f(y)|\leq M-m$, this gives $\sup\{|f(x)-f(y)|: a\leq x,y\leq b\}\leq M-m$.
On the other hand, still, $a\leq x,y\leq b$, we have $f(x)-f(y)\leq|f(x)-f(y)|$ and hence $f(x)\leq f(y)+\sup\{|f(x)-f(y)|: a\leq x,y\leq b\}$, this is true for all $a\leq x\leq b$, we have $M\leq f(y)+\sup\{|f(x)-f(y)|: a\leq x,y\leq b\}$, so $M-\sup\{|f(x)-f(y)|:a\leq x,y\leq b\}\leq f(y)$, this is true for all $a\leq x\leq b$, so $M-\sup\{|f(x)-f(y)|: a\leq x,y\leq b\}\leq m$, and hence $M-m\leq\sup\{|f(x)-f(y)|: a\leq x,y,\leq b\}$.