Strict (or lack therefore) concavity of square root function

Question

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.

Answer 1

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 $\sqrt{t}>t$.

Answer 2

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 $[0,\infty)$.