Convexity with setting $ \lambda=\frac{1}{2}$ implies continuity?

88 Views Asked by At

I'm not sure if there is already an answer to this particular question, but here it goes,

(Definition of convexity that is used here: Let $D \subset \mathbb{R}$ be an (infinite or finite) interval. A function $f:D\longrightarrow \mathbb{R}$ is called convex if for any $x_1,x_2 \in D$ and any $\lambda$ with $0<\lambda<1$,

$$f(\lambda x_1+(1-\lambda)x_2)\leq \lambda f(x_1) + (1-\lambda) f(x_2)$$ is valid )

The following was already proven:

In an interval $I \subset S $ continuous function $f:I \longrightarrow \mathbb{R}$ is then convex if and only if

$(1)$ $$f\bigg(\frac{x+y}{2}\bigg)\leq\frac{f(x)+f(y)}{2} $$ for any $x,y \in I$.

In the proof, the ( -> ) part is almost direct by setting $\lambda=\frac{1}{2}$ into the definition of convexity. But as for ( <- ) I had to use the continuous property of the function, but I wonder if the property $(1)$ stated above actually implies continuity so that the theorem above doesn't need to state the continuity property, as of:

In an interval $I \subset S $ function $f:I \longrightarrow \mathbb{R}$ is then convex if and only if

$(1)$ $$f\bigg(\frac{x+y}{2}\bigg)\leq\frac{f(x)+f(y)}{2} $$ for any $x,y \in I$.

I tried also to find counterexamples but to no avail, as it is for me rather intuitive that the property $(1)$ should imply continuity but I'm not sure, maybe the proof can be similar to the proof of "convexity implies continuity". I would love to see some counterexamples (assuming they exist) or perhaps the proof. Thank you very much in advance!