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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."?

Danny_Kim
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    $g$ solved from definition gives square root composition with $f$ and there is well known Convex Monotone superposition theorem - may be this make sense? – zkutch Jun 23 '20 at 03:21
  • @zkutch Okay. Consider $g(x) = h(f(x))$, where $h(x) = \sqrt{x}$. Then, $g''(x) = h''(f(x)) f'(x)^2 + h'(f(x)) f''(x)$. Since $f$ in convex (i.e., $f''>0$) and $h$ is a square root function (i.e., $h''<0$, and $h'>0$), we have $h''(f(x)) f'(x)^2<0$ and $h'(f(x)) f''(x)>0$. Thus, $g''(x)$ is sum of negative and positive ones, so I cannot guarantee $g''(x)>0$. Is there something wrong with me? – Danny_Kim Jun 23 '20 at 03:55
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    I think nothing wrong with your reasoning - only there that will be case where we have concavity. Original theorem prove even doesn't use derivative https://ljk.imag.fr/membres/Anatoli.Iouditski/cours/convex/chapitre_3.pdf . Now, sorry, I have no time, but later I'll try to find answer it terms of derivative too. – zkutch Jun 23 '20 at 04:18
  • @zkutch Thank you for your kind comments and a helpful link. – Danny_Kim Jun 23 '20 at 04:22
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    The statement is false. Take $g(x)=x^{2/3}$ which is clearly not convex. Its square, $f(x)=x^{4/3}$ is convex. – GReyes Jun 23 '20 at 04:23
  • Yes, I see. Outer function in theorem also needs to be convex. I'll try to find time and investigate theorem in conditions of only monotonic outer function. To try help you in little I'll calculate in answer second derivative. – zkutch Jun 23 '20 at 08:47
  • @Danny_Kim It is easy to prove the convexity of $g(x)$ by other methods, rather than "f is convex, which means that g is also convex." Can you provide some images from the paper? – River Li Jun 23 '20 at 16:14
  • @RiverLi I captured the image in the paper: https://i.stack.imgur.com/sVL8A.png. Surely, the convexity of $g(x)$ can be proven easily using second derivative (since it is a single-variable function). However, what I was really curious about was the basis for arguing that the $g(x)$ is convex because its square is convex in the process of proving. – Danny_Kim Jun 24 '20 at 00:51
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    @Danny_Kim The statement "$f$ is convex, then $\sqrt{f}$ is also convex" is not true, some condition(s) for $f$ should be added, e.g., $f$ is non-negative and quadratic (the case in the paper). – River Li Jun 24 '20 at 02:05

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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 $x<-a$.

zkutch
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  • Thank you very much for your hard work. What I was really curious about was the basis for arguing that the g(x) is convex because its square is convex in the process of proving. However, your answer also helped me a lot. – Danny_Kim Jun 24 '20 at 00:57
  • @Danny_Kim. One more, may be casuistic, consideration: you, sure, know $Epi (f)$ so called epigraph of function. What if authors take epigraph of reverse function as opposite set to initial one? On example. $x^{\alpha}$ is convex for $\alpha >1$ against Ox axis, but its reverse one $x^{\alpha}$ when $\alpha \in (0,1)$ is convex against Oy axis. Sorry if very fantastic version. – zkutch Jun 24 '20 at 07:06
  • Comments are not fully understood, but thank you for opening a new way of interpretation. – Danny_Kim Jun 24 '20 at 09:54
  • hm.., sorry, it's my "perfect" English, may be. Try in another words: if we stay on $Ox$ looking up to $Oy$, then convex is $x^2$, but when we stay on $Oy$ looking up to $Ox$, then convex is $\sqrt{x}$ – zkutch Jun 24 '20 at 10:02