TITLE: Coercivity vs boundedness of operator
QUESTION [9 upvotes]: The definition of coercivity and boundedness of a linear operator $L$ between two $B$ spaces looks similar: $\lVert Lx\lVert\geq M_1\lVert x\rVert$ and $\lVert Lx\rVert\leq M_2\lVert x\rVert$ for some constants $M_1$ and $M_2$. Thus in order to show the existence of a PDE $Lu=f$ one needs to show that it is coercive. However if my operator $L$ happen to be bounded and $M_2 \leq M_1$?
What is the intuition behind those two concepts because they are based on computation of the same quantities and comparing the two?

REPLY [5 votes]: The number $\inf_{x\neq 0}\frac{\lVert Lx\rVert}{\lVert L\rVert\cdot\lVert x\rVert}$ measure how injective $L$ is (when $L=0$ it's not well defined). 
Consider $R(x):=\lVert Lx\rVert$ for $x$ in the unit ball. Coercivity means that $R(x)\geq M_1>0$ for some constant $M_1$ and in particular $L$ is injective (but $L$ can be unbounded, for example if $B=\ell^2(\Bbb R)$ and $Le_n=ne_n$, $n\geq 1$).
A bounded operator doens't need to be coercive, for example $L\equiv 0$. 
To get an intuition, coercivity means that the vectors of the unit ball are map to a positive distance from $0$, and this distance is independent of the point. Boundedness measure how far form $0$ can be mapped the vectors of the unit ball.