It is known that if $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is uniformly convex, i.e. there is $\mu > 0$ such that $$ f(tx+(1-t)y) \leq tf(x) + (1-t)f(y) -\mu t \lVert y-x \rVert^2~~~ \forall (x, y) \in \mathbb{R}^n, t \in (0, 1), $$ then $$\lim_{\lVert x \rVert \rightarrow \infty} f(x) = \infty.$$ One proof of the latter result I can think of is the following: Rearrange: $$ \tfrac{f(tx+(1-t)y)-f(y)}{t} \leq \big(f(x)-f(y)\big) -\mu \lVert y - x \rVert^2 $$ Use that $f$ is locally Lipschitz and Rademacher's theorem (both nontrivial results) and assume that $f$ is differentiable in $y$. Let $t \searrow 0$: $$ \nabla f(y)^\top (x-y) \leq f(x) -f(y) -\mu \lVert y - x \rVert^2 $$ Rearrange and use Cauchy-Schwarz again to conclude: $$ f(x) \geq f(y) + \mu\lVert y - x\rVert^2 -\lVert \nabla f(y) \rVert \lVert x -y \rVert $$ So we get the desired limit as $\lVert x \rVert \rightarrow \infty$, because the quadratic term dominates.
My question is if the above if this is correct, but more importantly, if there is a more elementary proof of this that does not use advanced analysis results like Rademacher. Thank you in advance.
