Show that $f(x, y, z) = xy + z^2$ is continuous. $(\varepsilon, \delta)$

Question

Show that $f: \mathbb{R^3} \to \mathbb{R}$ $$f(x, y, z) = xy + z^2$$ is continuous.

One could just deduce that since it's a polynomial it's continuous, but how would I show this using $(\varepsilon, \delta)$? I'm not familiar on using the method with multivariate functions.

Answer 1

Note that if $(x,y,z),(x_0,y_0,z_0)\in\Bbb R^3$, then\begin{align}\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|&=|xy-x_0y_0+z^2-z_0^{\,2}|\\&\leqslant|xy-x_0y_0|+|z^2-z_0^{\,2}|\\&=\bigl|(x-x_0)y_0+(y-y_0)x_0+(x-x_0)(y-y_0)\bigr|+\\&\phantom{=}+\bigl|(z-z_0)(z+z_0)\bigr|\\&\leqslant|x-x_0||y_0|+|y-y_0||x_0|+|x-x_0||y-y_0|+\\&\phantom{=}+|z-z_0|\bigl(|z-z_0|+2|z_0|\bigr).\end{align}So, given $\varepsilon>0$, take $\delta>0$ such that:

• $\delta|y_0|<\dfrac\varepsilon5$;
• $\delta|x_0|<\dfrac\varepsilon5$;
• $\displaystyle\delta<\frac{\sqrt\varepsilon}{\sqrt5}$;
• $2\delta|z_0|<\dfrac\varepsilon5$.

Then, assuming that $|x-x_0|,|y-y_0|,|z-z_0|<\delta$, since $\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|$ is smaller than or equal to$$\overbrace{|x-x_0||y_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|y-y_0||x_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|x-x_0||y-y_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|z-z_0|^2}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{2|z-z_0||z_0|}^{\phantom{\varepsilon/5}<\varepsilon/5},$$you have$$\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|<\varepsilon.$$

Answer 2

Hint: Denote $\xi = (x,y,z)$ and $\alpha = (a,b,c)$. Then, \begin{align} |f(\xi) - f(\alpha)| &= |(xy+z^2) - (ab+c^2)|\\ &= |(x-a)y + a(y-b) + (z-c)(z+c)|\\ & \leq |y| |x-a| + |a||y-b| + |z+c| |z-c| \\ &\leq \left( |y| + |a| + |z+c|\right) \lVert\xi - \alpha \rVert \end{align} Now, if $\lVert \xi-\alpha\rVert < 1$, can you find an upper bound on the thing in brackets?

If this is still too far of a leap, I suggest you take a look at the proof of how we prove the sums and product of continuous functions is continuous. In particular look at the proof of why $t\mapsto t^2$ is continuous from single-variable analysis.

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Notice that the idea is to use basic algebra and "force" terms like $x-a$ to appear, because if this is small enough, you can make the entire thing small enough.