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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?

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    If $0 then for all $x\neq 0$ we have $\lVert Lx\rVert=\lVert x\rVert$ so $L$ is an isometry. The constant of coercivity gives a below bound of $\lVert Lx\rVert$ when $x$ is in the unit ball, whereas the norm of $\lVert L\rVert$ is an upper bound.2012-03-23
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    thanks for the reply. So, I understand that coercivity is essential for proving existence of a solution, but what is the use of operator being bounded? for example consider a laplacian operator from L2 to L2, that's unbounded, however the same operator from H2 to L2 is bounded. Can I infer about existence of a solution in either case?2012-03-23
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    Side remark: coercive is neither necessary nor sufficient for solving the equation $Lu = f$ where $u \in U$ and $f\in F$ and $U$, $F$ are Banach spaces and $L$ is a linear operator. Just as in the finite dimensional case the solvability of the equation in general requires that $L$ be invertible (and hence bijective). Coercivity implies injectivity, but not the other way around. And even with injectivity you still need to make sure that $L$ is surjective before you can guarantee the existence of a solution.2012-06-15

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