$ \max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} = \underset{x \in X}{\sup } \{ \max \{ f(x), g(x) \} \}$?

Question

Let $f,g: X \rightarrow [0, \infty )$ be bounded functions.

Does hold that

$$ \max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} = \underset{x \in X}{\sup } \{ \max \{ f(x), g(x) \} \}$$

?

My attempt

Let $x \in X$. Then:

$$\max \{ f(x) , g(x) \} \leq \max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} $$

Taking the sup, we have:

$$\underset{x \in X}{\sup } \{ \max \{ f(x), g(x) \} \} \leq \max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} $$

If $\max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} = \underset{x \in X}{\sup f(x) }$ $\bigg(\underset{x \in X}{\sup g(x) }\bigg)$, we have:

$$\max \{ \underset{x \in X}{\sup } f(x)\ , \underset{x \in X}{\sup } g(x) \} \leq \underset{x \in X}{\sup } \{ \max \{ f(x), g(x) \} \} $$

since $f(x) \leq \max \{ f(x), g(x) \} \}$ $\bigg(g(x) \leq \max \{ f(x), g(x) \} \} \bigg)$, $\forall x \in X$

Am I right?

Answer 1

The second part looks good, though I'd proceed slightly differently. We don't need to worry about which supremum (if either) is the greater. Instead, we note that for any $x_0\in X$ we have $$f(x_0)\leq\max\bigl\{f(x_0),g(x_0)\bigr\}\leq\sup_{x\in X}\left(\max\bigl\{f(x),g(x)\bigr\}\right).$$ Hence, we have $$\sup_{x\in X}f(x)\le\sup_{x\in X}\left(\max\bigl\{f(x),g(x)\bigr\}\right).$$ Similarly, $$\sup_{x\in X}g(x)\le\sup_{x\in X}\left(\max\bigl\{f(x),g(x)\bigr\}\right),$$ and so $$\max\left\{\sup_{x\in X}f(x),\sup_{x\in X}g(x)\right\}\le\sup_{x\in X}\left(\max\bigl\{f(x),g(x)\bigr\}\right).$$

The first part needs more justification, to my mind. You can proceed in a similar way to the second part to get a nice proof with less "hand-waving."