Continuity for a two-variable function from baby Rudin While I'm studying Baby Rudin's Exercise, I got some problems.
The Exercise $9.28$ says I need to check the continuity of the function $$\varphi(x,t)=\begin{cases}x&0\leq x\leq \sqrt{t} \\ -x+2\sqrt{t}& \sqrt{t}\leq x\leq 2\sqrt{t}\\ 0&\text{otherwise}\end{cases}$$ and $\varphi(x,-t)=-\varphi(x,t)$ for $t\geq0$.
Intuitively, I know the function is continuous on $\mathbb{R}^2$, but I can't show it easily. Even I tried to show the continuity by $\varepsilon$-$\delta$ argument, but the formula got complicated; there are too many cases.
Is there any simple way to show its continuity? I heard that by using pasting lemma it can be proved easily, but it is out of the scope of the class.
 A: Check left and right endpoints of each point of disconnect. You will want to show this for $t<0, t=0, t>0$. Here I'll show $t>0$.
$$\lim_{x\to\sqrt{t}^-} \varphi(x,t) = x\big|_{\sqrt{t}} = \sqrt{t}$$
$$\lim_{x\to\sqrt{t}^+} \varphi(x,t) = (-x+2\sqrt{t})\big|_{\sqrt{t}} = \sqrt{t}$$
$$\varphi(\sqrt{t},t) = \sqrt{t} $$
Second,
$$\lim_{x\to2\sqrt{t}^-} \varphi(x,t) = (-x+2\sqrt{t})\big|_{2\sqrt{t}} = 0$$
$$\lim_{x\to2\sqrt{t}^+} \varphi(x,t) = 0\big|_{2\sqrt{t}} = 0$$
$$\varphi(2\sqrt{t},t) = 0 $$
A: The function
$$\psi(\xi):=\max\bigl\{1-|1-\xi|,\ 0\bigr\}\qquad(-\infty<\xi<\infty)$$
draws the piecewise linear $\phi$ along horizontal lines. This $\psi$ is obviously continuous. We then have
$$\phi(x,t)={\rm sgn}(t)\sqrt{|t|}\ \psi\!\left({x\over\sqrt{|t|}}\right)\quad(t\ne0),\qquad\phi(x,0)=0\ .$$
It follows that $\phi$ is continuous in ${\mathbb R}^2\setminus\bigl\{(x,0)\bigm| x\in{\mathbb R}\bigr\}$.
For a point $(x_0,0)$, let an $\epsilon>0$ be given. The set $U:={\mathbb R}\,\times\,]{-\epsilon^2},\epsilon^2[\>$ then is a neighborhood of $(x_0,0)$, and $|\psi(\xi)|\leq1$ implies
$$\bigl|\phi(x,t)-\phi(x_0,0)\bigr|<\epsilon\qquad\forall\, (x,t)\in U\ .$$
