Let $\phi \in \mathcal{M}_{M,N}(\mathbb{R})$, $y \in \mathbb{R}^M$ and $f(x)=\frac{1}{2}||\phi x – y||^2$.
I want to show that $f$ is coercive if and only if $\phi$ is injective.
This is what I have done :
For the first direction. Suppose that $\phi$ is not injective so there exists $x_0 \neq 0$ such that $\phi(x_0)=0$. For all integers $n$ define $x_n = n \cdot x_0$.
We see that $\displaystyle\lim_{n \to +\infty} ||x_n|| = +\infty$ whereas $\displaystyle\lim_{n \to +\infty} f(x_n) = \frac{1}{2}||y||^2$. Thus $f$ is not coercive.
For the other direction, i'm having some trouble. I noticed that
$$
\frac{1}{2}||\phi x – y||^2 \geq \frac{1}{2}(||\phi x|| – ||y||)^2
$$
If I show that $\displaystyle\lim_{||x|| \to +\infty} ||\phi x|| = +\infty$ I could conclude but I can't get any further.
Any comments are appreciated, thanks.
Best Answer
Injectivity of $\phi$ implies that $$ \|\phi(x)\|\ge c\|x\| $$ for some positive $c$. That forces the right hand side of your last inequality to diverge to $\infty$ as $\|x\|\to\infty$. Hence the left hand side diverges to infinity as well, and then $$ \|\phi(x)\|\ge\|\phi(x)-y\|-\|y\|\to\infty. $$