Assume $f$ and $g$ are continuous at $x=a$. Prove $h=\max\{f,g\}$ is continuous at $x=a$. Assume $f$ and $g$ are continuous at $x=a$. Prove $h=\max\{f,g\}$ is continuous at $x=a$.
My solution:
When $f\ge g\Rightarrow h=\max\{f,g\}=f$ and since $f$ is continuous at $x=a$ so is $h$.
When $f<g\Rightarrow h=\max\{f,g\}=g$ and since $g$ is continuous at $x=a$ so is $h$.
Does this seem sufficient?
 A: No, not at all. By the same argument, if $f$ and $g$ are differentiable, then so is $\max\{f,g\}$. However, $\max\{x,-x\}=\lvert x\rvert$.
A: We always have $x_k \le y_k + |x_k - y_k| \le y_k +\|x-y\|_\infty$.
Hence $x_k \le \max_j y_j +\|x-y\|_\infty$ and so $\max_j x_k \le \max_j y_j +\|x-y\|_\infty$.
Reversing the roles of $x,y$ gives $|\max_k x_k - \max_k y_k| \le \|x-y\|_\infty$.
In particular, the $\max$ is Lipschitz with rank one.
Now consider the function $x \mapsto \max(f(x),g(x))$.
Alternative:
Suppose $f(a) > g(a)$. Then there is a neighbourhood $U$ of $a$ such that $f(x)>g(x)$
for $x\in U$. Hence $\max(f(x),g(x)) = f(x)$ for $x \in U$ and so $h$ is continuous.
The case $f(a) < g(a)$ is similar.
If $f(a) = g(a)$, let $\epsilon>0$ and choose a neighbourhood $U$ such that
$|f(x)-f(a)| < \epsilon$ and $|g(x)-g(a)| < \epsilon$ for $x \in U$.
Then
$-\epsilon+f(a) < f(x) < f(a) + \epsilon$ and $-\epsilon+f(a) < g(x) < f(a) + \epsilon$ for $x \in U$ and
so $-\epsilon+f(a) <\max(f(x),g(x)) < f(a) + \epsilon$ for $x \in U$.
Hence $| h(x)-h(a) | < \epsilon$.
A: At $x=a$, use the fact that 
$$\max{(f,g)}=\frac{1}{2}(f+g+|f-g|).$$
