Definitions of coercivity - functional analysis

Question

In my PDE courses I've come across two different definitions or coercivity of a functional $\mathit{F}: \mathit{H} \rightarrow \mathbb{R}$ where $H$ is a Hilbert space.

Definition 1: https://mathworld.wolfram.com/CoerciveFunctional.html

Definition 2: $\mathit{F}$ is called coercive, if for some $a \in \mathbb{R},$ the corresponding sublevel set is non-empty and bounded.

Question 1: Are these two definitions equivalent?

Question 2: I've seen another definition where the Hilbert space is $\mathbb{R}^n$ where coercivity is defined as the property in Definition 2, but holding for every $a \in \mathbb{R}.$ Why is this difference?

Answer 1

Regarding question 1, the two notions are not equivalent because the first functional is a bilinear form, so is not like in the second definition for $f:H \rightarrow \mathbb{R}$. The first notion is for $B: H \times H \rightarrow \mathbb{R}$ bilinear. If a bilinear form $B: H \times H \rightarrow \mathbb{R}$ is continuous e coercive $(B(u,u) \geq c \|u\|^2$ for all $u \in H$, then for every given linear and continuous functional $\ell: H \rightarrow \mathbb{R}$, the equation $B(u,v) = \ell(v),\;\forall v \in H$, has a unique solution in $H$. If the bilinear form is symmetric, then the unique solution also solves $(1/2)B(u,u) - \ell(u)=min!$ in $H$.

(the point being that a minimum problem appears this way).

Regarding definition 2, there is a third notion of coercive functional: $f(x)$ goes to infinity as $\|x\| \rightarrow +\infty$. It is a key assumption in a well-known theorem: a weakly sequentially lower semi-continuous and coercive (in this third sense) functional is bounded below on H and has a global minimum in H [This coerciveness 3 ensures that the infimum of $F$ on H is the same as the infimum of $F$ over a sufficiently large closed ball (a weakly sequentially compact set), and this fact allows one to use another result that establishes the previous conclusion under the assumptions of existing a (non-empty) subset of H weakly sequentially compact with F weakly sequentially lower semi-continuous.]

Back to definition 2, it shown at On coercivity and compactness that for $f:\mathbb{R}^{n} \rightarrow \mathbb{R}$ continuous, this third notion of coercitivity is equivalent to every sublevel set being compact. So, in $\mathbb{R}^n$, under the assumption of f continuous one has boundedness of every sublevel set [implies Def. 2] being equivalent to coercitivity 3.

However, it is known that (in $\mathbb{R}^{n}$) coercivity 3 is too strong and one can prove the existence of a global minimizer using Def. 2 (in the proof one needs only one such level subset being bounded (thus compact).)

Answer 2

I think that the usual definition for a function $F \colon X \to \mathbb R$ to be coercive is that $$\lim_{\|x\| \to \infty } F(x) =\infty .$$ Now, for (proper) convex lower semi-continuous functions, the following are equivalent:

1. $f$ is coercive.

2. $\exists a>0,\ b \in \mathbb R$ such that $f \ge a \|\cdot\| +b$.

3. $\liminf_{\|x\| \to \infty } f(x)/\|x\|>0$, where $$ \liminf_{\|x\| \to \infty } \frac{f(x)}{\|x\|} = \lim_{r\to\infty} \sup_{\|x\| \ge r} \frac{f(x)}{\|x\|}.$$

4. The sublevel sets of $f$ are bounded.

Here is a proof: a) $\implies$ b): We may assume that $f(0)=0$ (if not, look at $g=f-f(0)$). Pick $r>0$ so that $$ \|x \| \ge r \implies f(x) \ge 1. $$ Then $$ 1 \le f \big ( \frac{r}{\|x\|} x\big) \le \frac{\|x\|-r}{\|x\|} f(0) + \frac{r}{\|x\|} f(x) $$ so that $f(x) \ge \tfrac 1r \|x\|$ for $\|x\| \ge r$. Notice that $$ \inf_{\|x\| <r} f(x) > - \infty $$ for if not, then there exists $(x_n)$ in $rB_X$ such that $f(x_n) \to - \infty.$ By passing to a subsequence if necessary we may assume that $$ f(x_n) < -n^2 $$ so that $$ f(\tfrac 1n x_n) \le \tfrac 1n f(x_n) < -n \to - \infty. $$ But $\tfrac 1n x_n \to 0$ so that by lsc $$ \liminf_{n \to \infty} f(\tfrac 1n x_n) \ge 0. $$ Hence, there exists $M>0$ such that $$ \|x\|<r \implies f(x) > -M. $$ Pick $a=1/r$ and $b = -M-1<0$. Then for $\|x\| \ge r$ $$ f(x) \ge \tfrac 1r \|x\| \ge a\|x\| +b $$ and for $\|x\|<r$ $$ \tfrac 1r \|x\| -M -1 <1- f(x) -1 =f(x). $$ b) $\implies$ c): If $f\ge a\|\cdot\|+b$ then $$ \liminf_{\|x\| \to \infty } \frac{f(x)}{\|x\|}\ge a>0. $$ c) $ \implies $ d): Suppose that there exists $\lambda$ such that the sublevel set $(f\le \lambda )$ is unbounded. Then there exists $(x_n)$ in $(f \le \lambda) $ with $\|x_n\| \to \infty$. Since $f(x_n) \le \lambda $ we obtain $$ \liminf_{n\to \infty} \frac{f(x_n)}{\|x_n\|} \le 0 $$ but $$ \liminf_{n\to \infty} \frac{f(x_n)}{\|x_n\|} \ge \liminf_{\|x\| \to \infty } \frac{f(x)}{\|x\|}>0. $$ d) $\implies$ a): If $f$ is not coercive then there exist $(x_n) \subset X$, $\lambda \in \mathbb R$ such that $\|x_n\| \to \infty$ and $f(x_n) \le \lambda$. This is not possible since $(f\le \lambda)$ is bounded.