Let $X$ be a locally convex topological vector space, and let $C$ be a nonempty convex subset of $X$. A real-valued function $f: C \to \mathbb R$ is strictly convex if for all $\lambda \in (0,1)$ and distinct $x,y \in C$ $$f(\lambda x + (1 - \lambda)y) < \lambda f(x) + (1-\lambda) f(y).$$
Does a strictly convex, continuous, real-valued function exist on every nonempty convex $C$?
No, not necessarily. In particular, consider some infinite-dimensional normed linear space $X$, but equip $X$ with the weak topology. Suppose there was a weakly continuous, strictly convex function from $C = X$ to $\Bbb{R}$. By adding affine functions and translating the graph as necessary, we may assume without loss of generality that $f$ achieves a minimum of $0$ at $0 \in X$.
Since $f$ is weakly continuous, $f^{-1}(-\infty, 1) = f^{-1}[0, 1)$ is a weakly open set. Weakly open sets in $X$ must contain a finite-codimensional affine subspace, and since $X$ is infinite-dimensional, this subspace is non-trivial. Choose a line contained in this non-trivial affine subspace, and identify it with $\Bbb{R}$. The result is a strictly convex function $$g : \Bbb{R} \to \Bbb{R}$$ such that $g(x) \in [0, 1)$ for all $x \in \Bbb{R}$.
But, this is impossible! Pick two distinct points $x_1, x_2 \in \Bbb{R}$ such that $x_1 < x_2$. Then, if $x > x_2$, $$g(x_2) \le \frac{x_2 - x_1}{x - x_1}g(x) + \left(1 - \frac{x_2 - x_1}{x - x_1}\right)g(x_1).$$ Take the limit as $x \to \infty$, remembering that $g(x)$ is bounded, and we see that $g(x_2) \le g(x_1)$. But, on the other hand, if $x < x_1$, then similarly, $$g(x_1) \le \frac{x_2 - x_1}{x_2 - x}g(x) + \left(1 - \frac{x_2 - x_1}{x_2 - x}\right)g(x_2),$$ hence as $x \to -\infty$, $g(x_1) \le g(x_2)$. That is, $g(x_1) = g(x_2)$, hence $g$ is constant, contradicting $g$ being strictly convex.