Questions about coerciveness and convexity

Question

I just have a few yes/no questions, and would really appreciate if you could correct me where I am wrong, and for what fundamental flaw I have.

1. Would the set of coercive functions a linear space?

Would it be ill-mannered to take a set say $\{x^2,x^4,...,x^{2n+2}\}$ and argue this is a linear space?

2. Is the set of coercive functions a conical set?

My answer would be no, as conical sets are bounded above since a set $A$ is conical if whenever $x ∈ A, t>0$ ,$tx ∈ A$

3. Is the set of coercive functions a convex set? Yes, since they are all bounded below.

Let $f : \mathbb{R}^n → \mathbb{R}$ be coercive. Then which of the following are also coercive?

a)$f + g$ with $g$ bounded below

       Yes

b)$f + g$ with $g$ bounded above

       No

c) $f + g^{−1}$ with $g$ coercive and $g(x) \neq 0 \forall x$

       (!!!)No idea

Answer 1

I assume the following meaning of "coercive": An extended valued map $f : X \to [-\infty,+\infty]$ is coercive if whenever $\|x\|\to+\infty$, $f(x)\to+\infty$. Here, the set of functions is called $F$.

1. The set $C$ of coercive functions do not form a linear space, because if $f(x)$ is coercive, $-f(x)$ is not coercive. Moreover, we can come across expressions like $+\infty-\infty$, which is nonsense.

2. No, the set of coercive functions is not a conical set. A set $C$ is conical if for every $t\geq 0$, $f \in C$ implies $tf \in C$. Take $t=0$, which gives that the zero function must be in $C$. But the zero function is not coercive. Note that if we exclude $t=0$, then $C$ would be "conical" in this limited sense.

3. Yes, the set $C$ is a convex set. Boundedness below, however, has nothing to do with this fact. A function may be bounded below and not coercive. But take $t\in (0,1)$, and $f,g\in C$, and it is straightforward to show that if $\|x\|\to+\infty$ then $tf(x)+(1-t)g(x)\to+\infty$ as well. Hence $tf + (1-t)g$ is in $C$.

4. a) Yes, correct. b) Correct, $g$ may be unbounded below. c) If $g$ is coercive, then $g^{-1}(x)\to 0 $ as $\|x\|\to\infty$. Therefore $f(x)+g^{-1}(x)\to +\infty$.

Answer 2

The set of coercive functions is not a linear space, because $-f$ is not coercive, if $f$ is coercive. So one don't get inverse elements of addition.

No its not conical because $t \geq 0$ and not $t>0$ so you can choose t=0 and so $tf$ is not in the cone.

It is convex. Take the convex combination $\lambda f +(1-\lambda)g$. So the coefficients are positive or the convex combination yields f (resp. g) and so it is coercive.

a) is correct. g is bounded below and so it cannot change the coercivity. b) is not correct because take for example $f(x)=x^2, g(x)=x^3$. c) It is correct, because $g^{-1}(x)$ vanishes at infinity.

Answer 3

Ah sorry you're right. Replace g(x) by $g(x)=-x^4$. I understand the test in that way, that one has to consider $f+g$ if g is bounded from above. $f$ cannot be bounded from above by coerciveness.

Edit: I wanted to post it under your comment.