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For the square root function $g(x) = \sqrt{x}$, we have

\begin{align} g'(x) = 0.5x^{-0.5} \\ g''(x) = -0.25x^{-1.5} \\ \end{align}

So the second derivative is negative everywhere except if $x = 0$. So $g(x)$ is strictly concave if $x > 0$. But when $x = 0$, the second derivative is undefined. What can we say about the strict concavity or the square root function at $x = 0$?

It seems that we can't prove strict concavity using the second derivative, but using Jensen's inequality, I think we can fairly easily show it, though even with Jensen's, we would have to assume that $t \in (0, 1)$ instead of $t \in [0, 1]$, or we run into the same issue.

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    $\begingroup$ Concavity is a property of a function on an interval rather than at one point. $\endgroup$
    – fwd
    May 5, 2021 at 1:00
  • $\begingroup$ @fwd But what's the property of the function on the interval $[0, a]$ for some $a > 0$? We still have an undefined second derivative at the lower bound. $\endgroup$
    – David
    May 5, 2021 at 1:01
  • $\begingroup$ Since derivatives involve two sided limits, I'm fairly sure that the second derivative of the square root is undefined. $\endgroup$
    – Kyky
    May 5, 2021 at 1:12
  • $\begingroup$ @Kyky But in this case, at $x = 0$, would it make sense to do consider the limit of $0^-$? The square root function isn't even defined for $x < 0$, so I feel we should only consider the limit from $0^+$? $\endgroup$
    – David
    May 6, 2021 at 14:56

2 Answers 2

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Another view:

One definition of concavity is as follows:

Suppose $a < b$. The function $g$ is [strictly] concave on the interval $[a,b]$ if for all $x$ with $a < x < b$ we have: $$\frac {g(x) - g(a)}{x - a} > \frac{g(b) - g(a)}{b - a}.$$

For the function $g$ defined by $g(x) = \sqrt x$ and the interval $[0,b]$ we have for $0 < x < b$, \begin{align} \frac {g(x) - g(0)}{x - 0} &= \frac {\sqrt x}{x}, \\ &= \frac{1}{\sqrt x}, \end{align}

and similarly

\begin{align} \frac {g(b) - g(0)}{b - 0} &= \frac {\sqrt b}{b}, \\ &= \frac{1}{\sqrt b}. \end{align}

Comparing, we see that for all $0 < x < b$,

$$\frac{1}{\sqrt x} > \frac{1}{\sqrt b},$$ or $$\frac {g(x) - g(0)}{x - 0} > \frac {g(b) - g(0)}{b - 0}.$$

Thus, $g$ is strictly concave on $[0,b]$. As this is true for any $b > 0$, $g$ is strictly concave on $[0,\infty)$.

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hint

Let $ a>0$.

$f:x\mapsto \sqrt{x} $ is strictly concave at $ [0,a]$ because it is syrictly concave at $(0,a] $ and

$$(\forall y\in (0,a])\; (\forall t\in(0,1))\;\;$$ $$ f(ty)=\sqrt{ty}>t\sqrt{y}$$ since $\sqrt{t}>t$.

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  • $\begingroup$ So we can only conclude that it's concave, but not that it's strictly concave? $\endgroup$
    – David
    May 5, 2021 at 1:15
  • $\begingroup$ @David What is the definition of "strictly concave " $\endgroup$ May 5, 2021 at 1:16
  • $\begingroup$ I just used the definition here en.wikipedia.org/wiki/….. $\endgroup$
    – David
    May 5, 2021 at 1:17
  • $\begingroup$ @David Yes it is strictly concave at $[0,\infty)$. $\endgroup$ May 5, 2021 at 1:21

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