Let $E$ be a Banach space, and $U \subset E$ an open non empty set. If $f:U\rightarrow \mathbb{R}$ is a convex function, show that this limist exists: $$\partial _y^+f(x) = \underset{t \rightarrow 0^+ }{lim}\frac{f(x+ty)-f(x)}{t} $$ My approach was to define the function $T_f(x_1,x_2)= \frac{f(x_2)-f(x_1)}{x_2 - x_1}$. As $f$ is convex, $T_f$ increase by each of its variables. Thus we have: $$T_f(x-ty, x) \leq T_f(x, x+ty) \leq T_f(x+ty, x+y) $$
But this doesn't seem to be able to answer the question fully because the left inequality does exist, as a convex function has right derivative, but I seem to be unable to find an inequality that would show the existence of the left derivative.