Let $X=(X_n)_{n>0}$ be a submartingale. Show that if $\phi$ is convex and nondecreasing on $\mathbb{R}$ and if $\phi(X_n)$ is integrable for each $n$, then $Y_n=\phi(X_n)$ is also a submartingale.
MY SOLUTION
We have that:
$1)$ $\mathbb{E}(|X_n|) < \infty$, each $n$;
$2)$ $X_n$ is $\mathcal{F_n}$-measurable, each $n$;
$3)$ $\mathbb{E}(X_n|\mathcal{F}_m) \geq X_m$ a.s., each $m \leq n$;
We have to show that, given that:
$4)$ $\phi^{''} > 0$;
$5)$ $\phi^{'} \geq 0$, that is: if $x < y$, then $\phi(x) \leq \phi(y)$;
$6)$ $\mathbb{E}(\phi(X_n)) < \infty $;
then:
$1.1)$ $\mathbb{E}(Y_n=\phi(X_n)|\mathcal{F}_m) \geq Y_m = \phi(X_m)$ a.s., each $m \leq n$.
FIRST, since $\phi$ is nondecreasing (assumption $5)$), given assumption $3)$, for each $m\leq n$ and almost surely it holds that: \begin{equation} \phi(\mathbb{E}(X_n|\mathcal{F}_m)) \geq \phi(X_m) = Y_m \end{equation}
SECONDLY, since $\phi$ is convex (assumption $4)$), by Jensen's inequality it holds that: \begin{equation} \mathbb{E}(\phi(X_n)=Y_n|\mathcal{F}_m) \geq \phi(\mathbb{E}(X_n|\mathcal{F}_m)) \geq \phi(X_m) = Y_m \end{equation}
which exactly corresponds to $1.1)$.
Is the above reasoning correct?
Your argument works fine but your understanding of convexity is not correct. Convex functions are not defined through second derivatives. They are defined by the inequality $\phi (tx+(1-t)y) \leq t\phi (x)+(1-t)\phi (y)$ and a convex function need not even have first derivative at very point. However Jensen's inequality is valid for these functions even if they are not differentiable.