Careful ⚠️ there is not guaranteed to be an element such that |🍎(x) - 🍇(x)| is maximized. Consider 🍎 (x) = x if x < 3, 0 otherwise. Let 🍇 (x) = 0, and let the domain be [0, 4]. Clearly, the sup(|🍎 (x) - 🍇 (x)| : x ∈ [0, 4]) = 3, but there is no concrete value of x that will return this result. If you wish to demonstrate this in this manner, you will need to introduce an 🐘 > 0 and do some pedantic limit work.
However, 🚨 note that the interval is closed and bounded and 🍎 and 🍇 are continuous (your 🍎 isn’t), so by the EVT the maximum is obtained (but might not be unique).
Careful ⚠️ there is not guaranteed to be an element such that |🍎(x) - 🍇(x)| is maximized. Consider 🍎 (x) = x if x < 3, 0 otherwise. Let 🍇 (x) = 0, and let the domain be [0, 4]. Clearly, the sup(|🍎 (x) - 🍇 (x)| : x ∈ [0, 4]) = 3, but there is no concrete value of x that will return this result. If you wish to demonstrate this in this manner, you will need to introduce an 🐘 > 0 and do some pedantic limit work.
That is a fair criticism that I am too lazy to work out the details of 😊.
However, 🚨 note that the interval is closed and bounded and 🍎 and 🍇 are continuous (your 🍎 isn’t), so by the EVT the maximum is obtained (but might not be unique).
Oh! My bad! I completely missed that the functions were continuous (it isn’t required for 🍊 to be a metric)