Prove/disprove $2|xx_0|-1<(x-x_0)^2$

Question

I'm working on a epsilon-delta relation, so I'm stuck at $$|f(x,y)-f(x_0,y_0)|=|x^2+y^2-x_0^2-y_0^2|\leq|x^2+y^2|+|x_0^2+y_0^2|$$ and I need to find epsilon in terms of the usual distance between two points in $R^2$. Since I take from the inequality this result $$|x^2+y^2|+|x_0^2+y_0^2|=(x-x_0)^2+(y-y_0)^2+2xx_0+2yy_0$$ and I need to factorize $(x-x_0)^2+(y-y_0)^2$ from that expresion, what I want to try is to say something like $2|xx_0|-1<(x-x_0)^2$ so that I'll be able to factorize this term $2xx_0+2yy_0$

Answer 1

\begin{align*} |x^{2}+y^{2}-x_{0}^{2}-y_{0}^{2}|&=|x^{2}-x_{0}^{2}+y^{2}-y_{0}^{2}|\\ &\leq|x^{2}-x_{0}^{2}|+|y^{2}-y_{0}^{2}|\\ &=|x+x_{0}||x-x_{0}|+|y+y_{0}||y-y_{0}|, \end{align*} we now let $(x-x_{0})^{2}+(y-y_{0})^{2}<\delta^{2}$, where $\delta<\min\{1,[2(1+|x_{0}|+|y_{0}|)]^{-1}\epsilon\}$, then $|x-x_{0}|<\delta<1$ and so is $|y-y_{0}|<1$, and hence $|x+x_{0}|=|x-x_{0}+2x_{0}|\leq|x-x_{0}|+2|x_{0}|<1+2|x_{0}|$, similarly, $|y+y_{0}|<1+2|y_{0}|$, back to the inequality we have \begin{align*} |x^{2}+y^{2}-x_{0}^{2}-y_{0}^{2}|&\leq(1+2|x_{0}|)|x-x_{0}|+(1+2|y_{0}|)|y-y_{0}|\\ &<(1+2|x_{0}|)\delta+(1+2|y_{0}|)\delta\\ &=2(1+|x_{0}|+|y_{0}|)\delta\\ &<2(1+|x_{0}|+|y_{0}|)[2(1+|x_{0}|+|y_{0}|)]^{-1}\epsilon\\ &=\epsilon. \end{align*}

Answer 2

Here is a way just using norm properties. $\delta_{\varepsilon}$ will be found naturally at the end of the chain of inequalities.

Let $u=(x,y)$ and $u_0=(x_0,y_0)$.

Your function in question is $f(u) = ||u||^2$ where $||\cdot||$ is the Euclidian norm. $U_{\delta}(u_0)$ denotes the $\delta$-circle around $u_0$.

$$|f(u) - f(u_0)| = |\,||u||^2 - ||u_0||^2\,| = |\,||u|| - ||u_0||\,|(||u|| + ||u_0||) $$$$ \leq ||u - u_0||(||u - u_0|| + 2||u_0||) \stackrel{u \in U_{\delta}(u_0)}{\lt} \delta(\delta + 2||u_0||)\stackrel{\delta \leq 1}{\leq}\delta(1+2||u_0||)$$

Now, you see:

$$\forall \varepsilon > 0 \;\exists \delta_{\varepsilon}= \frac{\min{(1,\varepsilon)}}{1+2||u_0||} \mbox{ s.t. } |f(u) - f(u_0)|< \varepsilon \, \forall u \in U_{\delta_{\varepsilon}}(u_0)$$