Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ defined by $f(x,y,z)=x^2y+2xz^2$ is continuous at $(1,1,1)$

Question

Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ definded by $$f(x,y,z)=x^2y+2xz^2$$ is continuous at $(1,1,1)$.

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Let $\epsilon<0$ then,

$$ \begin{equation} \begin{split} |x^2y+2xz^2-3| & = |(x^2y-y)+(y-1)+(2xz^2-2x)+(2x-2)| \\ & \leq |x^2y-y|+|y-1|+|2xz^2-2x|+|2x-2| \\ & = |y||x+1||x-1|+|y-1|+2|x||z+1||z-1|+2|x-1| \end{split} \end{equation} $$

For single variable I know how to handle $|x+1||x-1|$ $(\text{Using }|x+1||x-1|<\epsilon \implies\delta=\min\{1,\frac\epsilon3\})$, but for multivariable how to avoid any dependencies between variables?

It will be great help If anyone provide a legit way/hint/idea to deal this kind of problem(more general will be appreciated).

Answer 1

Let shift the limit point in $(0,0,0)$ (i.e. $x=1+u$ with $u\to 0$ and so on).

$\begin{align}|f(x,y,z)-f(1,1,1)| &=|f(1+u,1+v,1+w)-3|\\ &=|v+4u+2uv+u^2+u^2v+4w+2w^2+4uw+2uw^2|\\ &\le|v|+4|u|+2|u||v|+|u|^2+|u|^2|v|+4|w|+2|w|^2+4|u||w|+2|u||w|^2\end{align}$

In $\mathbb R^3$ all norms are equivalent, in the present case $||(x,y,z)||_\infty=\max(|x|,|y|,|z|)$ is by far the most convenient to use.

Notice that $||(u,v,w)||_\infty<\delta$ means that:

• by a choice of $\delta=1$ we have $|u|<1\implies |u|^n<|u|<1$
• by a choice of $\delta=\epsilon$ we have $|u|<\epsilon$

Same for the other letters, therefore for $\delta=\min(1,\epsilon)$ then any block $$|u|^m|v|^n|w|^p\le\epsilon\times 1\times 1\le \epsilon$$

We can continue like below:

$$|f(x,y,z)-f(1,1,1)|\le \epsilon+4\epsilon+2\epsilon+\epsilon+\epsilon+4\epsilon+2\epsilon+4\epsilon+2\epsilon\le 21\epsilon$$

And we are done.

Ultimately you can select $\delta=\min(1,\frac{\epsilon}{21})$ to get to a bare epsilon in the end, but not really required.

The dependencies between variables is hidden mostly in the choice of a suitable norm, but also in ignoring terms of higher degree (instead of having $\epsilon^2,\epsilon^3$ you just have $<\epsilon$ by bounding the other variables by $1$).

This is why the schema $\delta=\min(1,\epsilon)$ is so frequently used.

Answer 2

Alternative approach

Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\rightarrow\mathbb{R}$ definded by $f(x,y,z)=x^2y+2xz^2$ is continuous at $(1,1,1)$.

Assume that $\delta > 0.$

If $|(x,y,z) - (1,1,1)| < \delta$, then you can assume that
$-\delta < x-1 < \delta$
$-\delta < y-1 < \delta$
$-\delta < z-1 < \delta$.

This means that you can assume that
$1 - \delta < x < 1 + \delta$
$1 - \delta < y < 1 + \delta$
$1 - \delta < z < 1 + \delta$.

Therefore, for $x,y,z$ that satisfy these constraints,
the minimum value of $x^2y$ will be $> (1 - \delta)^3$ and
the maximum value of $x^2y$ will be $< (1 + \delta)^3$

Therefore, the minimum and maximum values for $f(x)$ will be $3(1 - \delta)^3$ and $3(1 + \delta)^3.$

Therefore, you need to set $\delta$ in terms of $\epsilon$
so that $[3(1 - \delta)^3 - f(1,1,1)] > -\epsilon$ and $[3(1 + \delta)^3 - f(1,1,1)] < \epsilon$.

For convenience, I will arbitrarily insist that $\delta$ will be $\leq (1/10).$
This allows me to assume that $(1 - \delta)^3 > (1 - 4\delta)$
and that $(1 + \delta)^3 < (1 + 4\delta)$.

So, the problem has been reduced to specifying $\delta$ in terms of $\epsilon$
so that $[3(1 - 4\delta) - 3] > - \epsilon$ and $[3(1 + 4\delta) - 3] < \epsilon$.

These constraints are satisfied when $12\delta < \epsilon$.

So, take $\delta = \min[(\epsilon/15), (1/10)].$