An Inequality Involving $\min(x, y)$

Question

The following problem is from Spivak's Calculus (4th ed., pg. 18):

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Prove that if:

$|x-x_0| < \min(\frac{\epsilon}{2(|y_0|+1)}, 1)$ and $|y - y_0| < \frac{\epsilon}{2(|x_0|+1)}$ then $|xy-x_0 y_0| <\epsilon$.

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What I've done so far is to say that the since the first inequality implies $|x-x_0| < 1$, combining this with the second inequality yields $|(x-x_0)(y-y_0)|<\frac{\epsilon}{2(|x_0|+1)}$, or $|(x-x_0)(y-y_0)|\cdot(|x_0|+1) < \frac{\epsilon}{2}$. We also know from the first inequality in the problem that $|x-x_0|<\frac{\epsilon}{2(|y_0|+1)}$, so $|x-x_0|(|y_0|+1)< \frac{\epsilon}{2}$. Combining yields: $|(x-x_0)(y-y_0)| \cdot(|x_0|+1) +|x-x_0|(|y_0|+1) < \epsilon$. I've tried manipulating the left hand side to get $|xy-x_0y_0|$ with little success. Does this whole approach to the problem seem ill-conceived? And if so, how can I rectify it? Or should I try a totally different approach? Thanks.

Answer 1

My approach would have been:

$|xy-x_{0}y_{0}|=|(xy-xy_{0})+(xy_{0}-x_{0}y_{0})|\le|x||y-y_{0}|+|y_{0}||x-x_{0}|$

$<|x|\big(\frac{\epsilon}{2(|x_{0}|+1)}\big)+|y_{0}|\big(\frac{\epsilon}{2(|y_{0}|+1)}\big)=\frac{|x|}{|x_{0}|+1}\frac{\epsilon}{2}+\frac{|y_{0}|}{|y_{0}|+1}\frac{\epsilon}{2}<\frac{|x|}{|x_{0}|+1}\frac{\epsilon}{2}+\frac{\epsilon}{2}$

Then noting that $|x|-|x_{0}|\le|x-x_{0}|<1$ (using that $|x-x_{0}|<1$ since it is bounded by the min of $1$ and $\frac{\epsilon}{2(|x_{0}|+1)}$) so $|x|<|x_{0}|+1$. So the above is bounded by

$<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Answer 2

This is a standard trick to add and subtract a term so that you end up via triangle inequality with two quantities that you have bounds for:

The quantity you want to bound can be written

$|xy - x_0 y_0| = |xy - x_0y + x_0y - x_0 y_0 | \leq |xy - x_0 y| + |x_0y - x_0 y_0|$

to get you started. You will also need to do one more step to get a $y_0$ into the first piece.