Showing continuity of a max function

Question

Let $f$ and $g$ be continuous functions on $X$ where $(X,d)$ is a metric space.

Define $h(x)=\max\{f(x),g(x)\}$. Show that $h$ is continuous on $X$.

Let $\epsilon>0$ and $x_{0}\in\mathbb{R}$

Without loss of generality, suppose $f(x_{0})>g(x_{0})$

$\implies |h(x)-h(x_{0})|=|h(x)-f(x_{0})|$

Since $g$ is continuous on $X$,

$\exists\space\delta_{0}>0$ such that if $|x-x_{0}|<\delta_{0}$ then $|g(x)-g(x_{0})|<f(x_{0})-g(x_{0})$ which implies $g(x)<f(x_{0})$

Since $f$ is continuous on $X$,

$\exists\space\delta_{1}\space(0<\delta_{1}\le\delta_{0})$ such that if $|x-x_{0}|<\delta_{1}$ then $|f(x)-f(x_{0})|<f(x_{0})-g(x)$ which implies $f(x)>g(x)$

So let $\delta=\delta_{1}$

If $|x-x_{0}|<\delta$ then $|h(x)-f(x_{0})|=|f(x)-f(x_{0})|<\epsilon$ since $f$ is continuous on $X$

$\therefore\space h$ is continous on $X$

Answer 1

You can use the fact that $h(x)=\frac{f(x)+g(x)+|f(x)-g(x)|}{2}$.

Answer 2

No. That $|f(x)-f(x_{0})|<f(x_{0})-g(x)$ which $f(x_{0})-g(x)$ has the variable $x$.

One can prove this assertion by observing that \begin{align*} |h(x)-h(y)|&=|(f\vee g)(x)-(f\vee g)(y)|\\ &=|f(x)\vee g(x)-f(x)\vee g(y)+f(x)\vee g(y)-f(y)\vee g(y)|\\ &\leq|f(x)\vee g(x)-f(x)\vee g(y)|+|f(x)\vee g(y)-f(y)\vee g(y)|\\ &\leq|g(x)-g(y)|+|f(x)-f(y)|. \end{align*}

Answer 3

There are a couple of things that are not completely correct in your formulation of the proof. First $X$ is some abstract metric space, not necessarily $\mathbb{R}$. Second you need to use the metric function $d_X(\cdot,\cdot):X\times X\to \mathbb{R}^+$ and $d_Y(\cdot,\cdot):Y\times Y\to \mathbb{R}^+$ in your arguments ($f,g:X\to Y$). For example the fact that $f$ and $g$ are assumed to be continuous in $X$ means that for any $x_0\in X$ you pick, for every $\varepsilon>0$ there are $\delta_1,\delta_2>0$ such that $$d_X(x,x_0)<\delta_1\Rightarrow d_Y(f(x),f(x_0))<\varepsilon\\ d_X(x,x_0)<\delta_1\Rightarrow d_Y(g(x),g(x_0))<\varepsilon$$ Now what you need to show is that $h(x):=\max\{f(x),g(x)\}$ is continuous under the above assumptions. That means for every $x_0\in X$ and for every $\varepsilon>0$ there is $\delta>0$ such that $$d_X(x,x_0)<\delta\Rightarrow d_Y(h(x),h(x_0))<\varepsilon$$ Now suppose at $x_0$ we have $f(x_0)> g(x_0)$ (the other case is analogue). Then by triangle inequality $$d_Y(h(x),h(x_0))=d_Y(h(x),f(x_0))\leqslant d_Y(h(x),f(x))+d_Y(f(x),f(x_0))$$ Since $f(x), g(x)$ are continuous and by assumption $f(x_0)>g(x_0)$ then there exists an open ball $B_{\delta}(x_0)$ where $ f(x)\geqslant g(x)$. Pick $\delta\leqslant \min\{\delta_1,\delta_2\}$ then $$d_Y(h(x),f(x))+d_Y(f(x),f(x_0))=d_Y(f(x),f(x))+d_Y(f(x),f(x_0))=d_Y(f(x),f(x_0))<\varepsilon$$ whenever $d_X(x,x_0)<\delta$ i.e. whenever $x\in B_{\delta}(x_0)$. Hence $h(x)$ is a continuous function in $X$.