Maximum and Supremum

Question

Please, I want to know

$\max_{x \in \{0,\frac{3}{4},1\}} (x-1/2)^2$, $\max_{x \in [0,1)} \min\{x,1/2\}$ and their $\arg \max$.

Thanks!

Answer 1

For the first part

$max_{x\in\{0,\frac34, 1\}}(x-\frac12)^2$=$max\{\frac14, \frac1{16}, \frac14\}$=$\frac 14$ and the argmax are $0,1$.

Note you could have easily predicted the max and the argmax by observing that the function to be maximised represents a parabola with vertex at $(\frac 12,0) $ with axis parallel to $y-axis$(since the parabolic function is decreasing on left and increasing on right of vertex)

Now for the second part

Take $A=[0,\frac12]$ and $B=(\frac 12,1)$

Then for $x \in A$ , $min\{x,\frac12\}=x$ and for $x\in B$, $\min \{x,\frac12\}=\frac12$

So $max_{x\in[0,1)}min\{x,\frac12\}=max\{\min_{x\in A}\{x,\frac12\}\cup min_{x\in B}\{x,\frac12\}\}$

$=max\{A\cup\{\frac12\}\}=\frac12$

Clearly the argmax is $B\cup \{\frac12\}=[\frac12,1)$

Hope this helps!