# Strict (or lack therefore) concavity of square root function

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.

• Concavity is a property of a function on an interval rather than at one point.
– fwd
May 5, 2021 at 1:00
• @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. May 5, 2021 at 1:01
• Since derivatives involve two sided limits, I'm fairly sure that the second derivative of the square root is undefined.
– Kyky
May 5, 2021 at 1:12
• @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 $0^+$? May 6, 2021 at 14:56

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 $$[0,\infty)$$.

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 $$\sqrt{t}>t$$.

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