Could someone verify the following absolute value inequalitiy:
$b_k < |\epsilon + L| \iff b_k < \epsilon + L $ and $bk < -(\epsilon+L) \iff -bk > (\epsilon +L)$
All together:
$-b_k > (\epsilon +L) > b_k$
Is there any further way to simplify this?
Could someone verify the following absolute value inequalitiy:
$b_k < |\epsilon + L| \iff b_k < \epsilon + L $ and $bk < -(\epsilon+L) \iff -bk > (\epsilon +L)$
All together:
$-b_k > (\epsilon +L) > b_k$
Is there any further way to simplify this?
I think it is correct. Further simplification would be dependent on whether you are seeking bounds for $L$ or for $\epsilon$:
You have $bk < (L+\epsilon) < -bk$, which is equivalent to