convex functions, is it enough to check if convex on all intervals

Question

Let $f: X \to \mathbb{R}$ be some function defined on a convex set $X \subset \mathbb{R}^n$. Now assume that for all $x \in X$ and $y \in \mathbb{R}^n$ such that $x+ty \in X$ for all $t \in [0,1]$, the function $f(x+ty)$ is a convex function on $[0,1]$. Does that mean $f$ is convex? To me the standard definition of convexity reduces to the case of intervals anyway, so intuitively my statement seems equivalent to $f$ being convex, but I don't see an argument why the definitions are equivalent.

Answer 1

Yes, it does. The usual definition is that $f$ is convex if for all $u, v \in X$ and $0 \le s \le 1$, $$f(su + (1-s) v) \le s f(u) + (1-s) f(v)$$ That is true (for any given $u$ and $v$) if the function $t \mapsto f(u + t (v - u))$ is convex on $[0,1]$.