Exercise 2.3.6

Note that to deduce the second part we can: First let $a$ denote the addition function, and $s$ denote the subtraction function, then note that $f+g=a\circ (f\oplus g)$ and $f-g=s\circ (f\oplus g)$. Now using exercises 2.3.4 and 2.3.5 we immediately get that $f+g,f-g$ are both uniformly continuous.

We will, however, just prove the generalization (second part) which automatically gives us the first.

(a)

$f+g$ is uniformly continuous.

Proof. Since $f,g$ are uniformly continuous we know that given ${𝜖}^{\prime }>0$ there is ${\delta }^{\prime }>0$ such that if $d_X\left(x,x_0\right)<{\delta }^{\prime }$ then $d_{ℝ}\left(f(x),f\left(x_0\right)\right)<{𝜖}^{\prime }$ and $d_{ℝ}\left(g(x),g\left(x_0\right)\right)<{𝜖}^{\prime }$ for all $x\in X$.

Given $𝜖>0$ let ${𝜖}^{\prime }=𝜖∕2$ and take $\delta ={\delta }^{\prime }$.

If $d_X\left(x,x_0\right)<\delta ={\delta }^{\prime }$ then from the uniform continuity of $f,g$ we know that $d_{ℝ}\left(f(x),f\left(x_0\right)\right)=$ $\left|f(x)-f\left(x_0\right)\right|<{𝜖}^{\prime }=𝜖∕2$ and $d_{ℝ}\left(g(x),g\left(x_0\right)\right)=\left|g(x)-g\left(x_0\right)\right|<{𝜖}^{\prime }=𝜖∕2$ But,

$$\begin{array}{ccc}\hfill \left|(f+g)(x)-(f+g)\left(x_0\right)\right|& =\left|f(x)+g(x)-\left(f\left(x_0\right)+g\left(x_0\right)\right)\right|\hfill & \hfill \\ \hfill & =\left|f(x)-f\left(x_0\right)+g(x)-g\left(x_0\right)\right|\hfill \\ \hfill & \le \left|f(x)-f\left(x_0\right)\right|+\left|g(x)-g\left(x_0\right)\right|\hfill \\ \hfill & <𝜖∕2+𝜖∕2\hfill \\ \hfill & =𝜖\hfill \end{array}$$

Thus, $f+g$ is uniformly continuous. □

(b)

$f-g$ is uniformly continuous.
A similar argument shows that $f-g$ is uniformly continuous.

(c)

$fg$ is not uniformly continuous.

Proof. To show that $fg$ is not uniformly continuous we take $f:(0,\infty )\to ℝ,g:(0,\infty )\to ℝ$ defined by $f(x)=g(x)=x$. So $fg=x^2$ which we know is not uniformly continuous. □

(d)

$\max <!--FUNCTION\; APPLICATION-->\{f,g\}$ is uniformly continuous.

Proof. Suppose $f,g$ are both uniformly continuous. That is, given ${𝜖}^{\prime }>0$ there exists ${\delta }^{\prime }>0$ such that if $d_X\left(x,x_0\right)<{\delta }^{\prime }$ then $\left|f(x)-f\left(x_0\right)\right|<{𝜖}^{\prime }$ and $\left|g(x)-g\left(x_0\right)\right|<{𝜖}^{\prime }$ for all $x\in X$. (which we can do if we just take the smaller of the two deltas that we would get from the original definition).

Now we want to show that given $𝜖>0$ there exists $\delta >0$ such that if $d_X\left(x,x_0\right)<\delta$ then $\left|\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))-\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)\right|<𝜖$ for all $x\in X.$

So let us break this into four cases.

1: $\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))=f(x),\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)=f\left(x_0\right)$. Given $𝜖>0$ take ${𝜖}^{\prime }=𝜖$ and $\delta ={\delta }^{\prime }$ then we immediately get that

$$\left|\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))-\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)\right|<𝜖\text{. }$$Case 2: $\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))=g(x),\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)=g\left(x_0\right)$ This case is exactly the same as Case $1.$ Case 3: $\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))=f(x),\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)=g\left(x_0\right)$. This case is a bit more complicated. So we have $f(x)>g(x)$ and $g\left(x_0\right)>f\left(x_0\right)$. And we can assume that $f(x)\ne g\left(x_0\right)$ (if $f(x)=g\left(x_0\right)$ then we are clearly done). Without loss of generality we can assume that $x<x_0$. Consider $h=f-g$ on the interval $\left[x,x_0\right]$. Then $h(x)>0$ and $h\left(x_0\right)<0$. And since both $f,g$ are continuous we know that $h$ is continuous and hence by the Intermediate Value Theorem we know that there exists $z\in$ $\left[x,x_0\right]$ such that $h(z)=0$. That is $f(z)=g(z)$. Now given $𝜖>0$ take ${𝜖}^{\prime }=𝜖∕2$ and take $\delta ={\delta }^{\prime }$ and we get: $$\begin{array}{ccc}\hfill \left|\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))-\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)\right|& =\left|f(x)-g\left(x_0\right)\right|\hfill & \hfill \\ \hfill & =\left|f(x)-g\left(x_0\right)-h(z)\right|\text{ since }h(z)=0\hfill \\ \hfill & =\left|f(x)-f(z)+g(z)-g\left(x_0\right)\right|\hfill \\ \hfill & \le |f(x)-f(z)|+\left|g(z)-g\left(x_0\right)\right|\hfill \\ \hfill & \le 𝜖∕2+𝜖∕2\hfill \\ \hfill & =𝜖\hfill \end{array}$$ as desired. Case 4: $\max <!--FUNCTION\; APPLICATION-->(f(x),g(x))=g(x)$, and $\max <!--FUNCTION\; APPLICATION-->\left(f\left(x_0\right),g\left(x_0\right)\right)=f\left(x_0\right)$. This is exactly the same as case $3.$

Therefore $\max <!--FUNCTION\; APPLICATION-->(f,g)$ is uniformly continuous whenever $f,g$ are uniformly continuous. □

•••(e)

$\min <!--FUNCTION\; APPLICATION-->\{f,g\}$ is uniformly continuous.
A similar argument shows that $\min <!--FUNCTION\; APPLICATION-->\{f,g\}$ is uniformly continuous.

(f)

$f∕g$ is not uniformly continuous.

Proof. Consider $f=x,g=1∕x$ where $f,g:(1,\infty )\to ℝ$. Then $f∕g=x^2$ which is not uniformly continuous. □

(g)

$cf$ is uniformly continuous.
Obvious.