I am reading a paper, in which the authors said that
For given $0<\theta_{min}<\theta<\frac{\pi}{2}$, $h>0$, and $d>0$, a function $f(x)$ is defined as $$f(x)=g(x)^2=d^2+h^2+x^2-2dx\cos(\theta)+2hx\sin(\theta).$$ Since the second derivative of $f$ with respect to $x$ is positive, $f$ is convex. It means that $g(x)$ is also convex with respect to $x$.
Is the above statement correct? I think that the state ment, if $\frac{d^2f(x)}{dx^2}>0$, then $g(x)$ such that $f(x)= g(x)^2$ is convex, is incorrect generally.
Even, I can provide counterexample as follows: $g(x) = |x|^{0.5}$. In this case, $f(x)$ will be $|x|$. Then, $g(x)$ is not convex, but $f(x)$ is convex.
Can someone explain why and how the authors said that "$f$ is convex, which means that $g$ is also convex."?
May be, this will be more helpful then general considerations. If we take $g(x)=\sqrt{x^2+ax+b}$, then $$g^{''}(x)=\frac{x(2a-2b)+4b-ba}{4\sqrt[\frac{3}{2}]{x^2+ax+b}}$$ so it's behavior is heavily depends on coefficients and is not convex always. At end let me share one notice, that if $g(x)=\sqrt{x^2+ax}$, then second derivative is non comparable more easy $$g^{''}(x)=\frac{-a^2}{4(x^2+ax)\sqrt{x^2+ax}}$$
So you need only $x<-a$.