For a convex function, does having a unique minimizer imply that it is coercive?

Question

I am looking for a simple proof that if a functional is convex and it has a unique minimizer, then it must be coercive. To be specific, let ${\cal H}$ is a Hilbert space and $G$ is a functional on it. If $G$ has a unique minimizer on ${\cal H}$, i.e. $\exists f_0 \in {\cal H}$ such that $G(f_0) < G(f), \quad \forall f \in {\cal H}, f \neq f_0$. How can we prove that $G(f) \rightarrow \infty$ when $\|f\| \rightarrow \infty$.

Answer 1

It is not true: take $\ell_2$, and let $e_n$ be a Hilbert basis. Then $$G(f)=\sum_n \frac1n \langle f,e_n \rangle^2$$ is convex, its unique minimum is zero, but for instance $G$ is bounded along the sequence $\sqrt n e_n$, which goes to infinity in norm.

EDIT: if the dimension is finite, it is true. Call $N$ the dimension. Suppose without loss of generality that $G(0)=0$ is the minimum. Since $G$ is convex, then it is continuous. Therefore its minimum $m$ on $\mathbb S^{N-1}$ is strictly positive, by compactness and the strict inequality for the minimum. Then by convexity, along every half line through the origin the growth of $G$ is at least linear with slope $m$, i.e. $ G(f)\geq m \|f\|$, from which the result follows.