how to show $f$ attains a minimum?

Question

Let $H$ be a Hilbert space and let $f\colon H\rightarrow \mathbb{R}$ be a continuous convex function such that $f(x_n)\rightarrow\infty$ whenever $\lVert x_n\rVert\to\infty$. We need to show that $f$ attains a minimum. I really need answer for this one, failed to do the problem, will appreciate for help.

Answer 1

It is straightforward to show that the level sets of $f$ are bounded (ie, $\forall \bar f, \exists M$ such that $\{ x | f(x) \leq \bar f\} \subset \bar B(0,M)$). Thus for some $M$, we have $\inf_{x \in H} f(x) = \inf_{||x|| \leq M} f(x)$.

Now consider the 'extended' Hilbert space $H \times \mathbb{R}$ with the obvious inner product, and look at the 'clipped' epigraph of $f$, $E = \{(x,\alpha) \; | \; ||x|| \leq M, f(x) \leq \alpha \}$. It should be clear that $E \subset H \times \mathbb{R}$ is closed and convex, and that the '$x$' component is bounded.

Define a functional on $E$ by $\phi((x,\alpha)) = \alpha$, and note that $\inf_{||x|| \leq M} f(x) = \inf_{(x,\alpha) \in E} \phi((x,\alpha))$. Let $(x_n, \alpha_n)$ be a minimizing sequence, ie, $\lim_{n \rightarrow \infty} \phi((x_n,\alpha_n)) = \inf_{(x,\alpha) \in E} \phi((x,\alpha))$. Since $||x_n|| \leq M$, and the closed ball is weakly compact, we have $x_n \rightarrow \hat x$ weakly, for some $\hat x \in H$ (in fact, $||\hat x|| \leq M$). Since we can choose $\alpha_n \rightarrow \hat \alpha$ (it is a minimizing sequence), we see that $(x_n,\alpha_n) \rightarrow (\hat x,\hat \alpha)$ weakly in the extended space. Consequently, $\phi((x_n,\alpha_n)) \rightarrow \phi((\hat x,\hat \alpha))$, since $\phi$ is bounded.

It should be clear that $\hat \alpha = f(\hat x)$ (otherwise a quick contradiction), and whenever $||x|| \leq M$, we have $(x,f(x)) \in E$. Consequently we have $f(\hat x) \leq f(x)$, whenever $||x|| \leq M$, and by choice of $M$, it follows that $\hat x$ minimizes $f$ everywhere.