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The following question arises from another exercise of Ransford's book.

Let $-\infty\leq a<b\leq+\infty$ and $f\colon(a,b)\rightarrow\mathbb{R}$. We say that $f$ is convex if for $$c=(1-\lambda)x+\lambda y\Rightarrow f(c)\leq (1-\lambda)f(x)+\lambda f(y)\qquad (0\leq \lambda\leq 1.$$

It is true that every convex function defined on $\mathbb{R}$, which is bounded above must be constant. This statement has been dealt with extensively and answered, for instance, in here. Nevertheless, the question we ask in the book is the following one.

Use this result to prove the Liouville Theorem for subharmonic functions, that is to prove that $$\text{every subharmonic function on }\mathbb{C}\text{ which is bounded above is constant}.$$ Information about subharmonic functions can be found in another related questions. However, the only result that is displayed in the aforementioned book that connects, in a sense, convex and subharmonic functions, is the following.

Let $-\infty\leq a<b\leq+\infty$, $u\colon U\rightarrow [a,b)$ be a subharmonic function on an open set $U\subset\mathbb{C}$ and let $\psi\colon (a,b)\rightarrow\mathbb{R}$ be an increasing convex function. Then $$\psi\circ u\text{ is subharmonic on }U,\text{ where }\psi(a)\doteqdot\lim_{t\to a}\psi(t).$$

It is not clear to me if we can use this result to prove the Theorem of discussion. Any ideas discussed here will be of pleasure to me.

Best regards.

SprtWhitebeard
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    Show that subharmonic functions restricted to lines are convex so constant in this case and any two points are joined by a line (if $u$ is twice differentiable the result is obvious by computing the Laplacian as a positive multiple of the second derivative of the restriction to the line and in general one can use approximations by differentiable subharmonic functions) – Conrad Sep 26 '23 at 13:06
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    @Conrad: “subharmonic functions restricted to lines are convex” – Is that really so? $v(x, y) = x^2-y^2$ is (sub)harmonic, but the restriction to the imaginary axis $v(0, y) = -y^2$ is not convex. Perhaps I am misunderstanding something? – Martin R Sep 29 '23 at 08:26
  • Yes Martin R, your statement is true. It is also supported by the first exercise of 2.6 section of Ransford, which challenges you to show that there exist subharmonic functions such that restricted to horizontal or vertival lines are not convex. – SprtWhitebeard Oct 05 '23 at 09:06

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