Concavity of a function

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Concavity of a function

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#1 h Yesterday at 9:00 AM

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Hi everyone,

I am studying the concavity of the function

$f(x) = \sqrt{1 - x^a}, \quad a \geq 0$
on the interval $x \in [0,1]$ .

I computed the second derivative and found that for $a \geq 1$ , the function appears to be concave. However, I am uncertain about the behavior at the endpoints.

Does anyone have insights on confirming concavity rigorously for $a \geq 1$ and understanding the behavior at the endpoints? Any help would be greatly appreciated!

Thanks!

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#2 h Yesterday at 1:50 PM

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We have $f'(x)={{-ax^{a-1}}\over {2\sqrt{1-x^a}}}$ . Now $x^{a-1}$ increases if $a>1$ and $1-x^a$ decreases. Hence ${1\over {\sqrt{1-x^a}}}$ increases and $f$ is indeed concave.
The endpoints are not a problem: if $f$ is concave on $(a,b)$ and continuous on $[a,b]$ , then it is also concave on $[a,b]$ .
The proof is easy: suppose $c<b, \lambda,\mu\ge 0, \lambda+\mu=1$ . Also choose $c<d<b$ . Then $f(\lambda c+\mu d)\ge \lambda f(c)+\mu f(d)$ .
Now send $d\uparrow b$ to obtain $f(\lambda c+\mu b)\ge \lambda f(c)+\mu f(b)$ . Same thing when the left point is $x=a$ .
And welcome to AOPS.

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