(Proof) If $f$ and $g$ are continuous, then $\max\{f(x),g(x)\}$ is continuous

Question

Consider the continuous functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$.

Show that $F:\mathbb{R}\rightarrow\mathbb{R}$ with $x\mapsto \max\{f(x),g(x)\}$ is continuous using the $\epsilon - \delta$ definition of continuity.

I know there must be four cases.

If $f(x)\leq g(x)$ and $f(x_0)\leq g(x_0)$ or

if $g(x)\leq f(x)$ and $g(x_0)\leq f(x_0)$ it is easy.

However, assuming $f(x_0)\neq g(x_0)$, what if

$g(x)\leq f(x)$ and $f(x_0)\leq g(x_0)$ or

$f(x)\leq g(x)$ and $g(x_0)\leq f(x_0)$?

For example:

$|f(x)-g(x_0)|$... how do I get from here to $|x-x_0|$?

Answer 1

Hint: The following identity may make the calculation more familiar. $$\max(a,b)=\frac{1}{2}\left(a+b+|a-b|\right).$$

Answer 2

First verify that for general $a,b,c,d\in\mathbb R:$

$$|\max(a,b)-\max(c,d)|<|a-c|+|b-d|.$$

Then apply that to show that, given explicit deltas for $f$ and $g$ separately that satisfy half a given epsilon, the appropriate delta for $\max(f,g)$ is the smallest of the individual deltas.

In general, if $f$ and $g$ are continuous on the same domain, then their cartesian product $f\times g$ is a continuous mapping from that domain to the cartesian product of the original two target sets.

The function "max" from $\mathbb R^2$ to $\mathbb R$ is continuous.

Your function is then the composition of two continuous mappings.

Answer 3

Given $\epsilon>0$, find $\delta_f$ so that when $|x-x_0|<\delta_f$, $|f(x)-f(x_0)|<\epsilon$. Similarly find $\delta_g$ for $g$.

Then define $\delta = \min(\delta_f,\delta_g)$. Prove this is good enough for $\max(f,g)$.

Answer 4

Trying to stick as close as possible to the definition of the maximum.

Take a point $x_0 \in \mathbb R$. You want to prove that $h(x)=\max (f(x),g(x))$ is continuous at $x_0$.

In fact you only have 3 cases:

1. $f(x_0) < g(x_0)$
2. $f(x_0)=g(x_0)$
3. $f(x_0) > g(x_0)$

$x$ will come into play afterwards.

Case 1. As $g(x_0)-f(x_0) > 0$, you can find $\delta$ such that for $\vert x - x_0 \vert < \delta$ you have $\vert g(x) - g(x_0) \vert < \frac{g(x_0)-f(x_0)}{2}$. Hence for $x \in (x_0-\delta,x_0+\delta)$ you have $h(x)=g(x)$ and $h$ is therefore continuous at $x_0$.

Case 3. Is similar to case 1.

Case 2. $f(x_0)=g(x_0)$

For $\epsilon > 0$, you can find $\delta_f$ such that $\vert f(x)-f(x_0) \vert < \epsilon$ for $\vert x - x_0 \vert < \delta_f$ and $\delta_g$ such that $\vert g(x)-g(x_0) \vert < \epsilon$ for $\vert x - x_0 \vert < \delta_g$.

Now for $\vert x- x_0 \vert < \inf(\delta_f,\delta_g)$, you have $f(x),g(x) \in (f(x_0)-\epsilon,f(x_0)+\epsilon)=(g(x_0)-\epsilon,g(x_0)+\epsilon)$ hence $h(x)=\max(f(x),g(x)) \in (f(x_0)-\epsilon,f(x_0)+\epsilon)$ which allows to conclude.