In the proof of this statement, we need to prove that $f$ is continuous at an arbitrary $x_0\in(a,b)$. My professor has "shifted the interval $(a,b)$" so that we can set $x_0=0$. We then fix $c>0$ such that both $c, -c\in(a,b)$.
From the definition of convexity, we write $$\theta c=\theta c+(1-\theta)0$$ so we have $$f(\theta c)\le\theta f(c)+(1-\theta)f(0).~~~~(1)$$
My professor then claims that if we take $\limsup$ as $\theta\to0$, we get $$\limsup_{x\to0^+}f(x)\le f(0).~~~~(2)$$
Similarly, from writing $$0=\theta(-c)+(1-\theta)\epsilon,$$ where $$\epsilon=\theta\frac{c}{1-\theta}.$$ Then from convexity we have $$f(0)\le \theta f(-c)+(1-\theta)f(\epsilon).~~~~(3)$$ Taking $\liminf$ as $\theta\to0$ we get $$f(0)\le \liminf_{x\to0^+}f(x).~~~~(4)$$
I understand how to use $(2)$ and $(4)$ to conclude that $f(0)=\lim_{x\to0^+}f(x)$ and similarly for $f(0)=\lim_{x\to0^-}f(x)$, but I don't understand how we go from $(1)$ to $(2)$ and from $(3)$ to $(4)$. What does $\limsup$ do for us in $(2)$ that $\lim$ doesn't, because to me, the right hand side of $(2)$ looks precisely like $$\lim_{\theta\to0^+}(\theta f(c)+(1-\theta)f(0)).$$ So can someone explain what it means to be taking $\limsup$ of $(1)$ and $\liminf$ of and $(3)$?
If $\epsilon >0$ the $\theta f(c)+(1-\theta)f(0) <f(0)+\epsilon$ provided $|\theta| <\frac {\epsilon} {|f(c)+|f(0)|}$. Hence, for any such $\theta$ we get $ f(\theta c)\leq f(0)+\epsilon$. Replacing $\theta$ by $\frac {\theta } c$ this says $f(\theta) <f(0)+\epsilon$ whenever $|\theta| < \frac {\epsilon |c|} {|f(c)+|f(0)|}$. Can use definition of $\lim \sup$ now to conclude that $\lim \sup_{\theta \to 0} f(\theta) \leq f(0)$?
[(3) implies (4) is proved in a similar way].