# Coercivity vs boundedness of operator

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 $0<M_2\leq M_1$ 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. – Davide Giraudo Mar 23 '12 at 22:35
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? – Medan Mar 23 '12 at 23:06
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. – Willie Wong Jun 15 '12 at 11:58

With boundedness everything is clear, because it is well known that linear operator $L$ is continuous iff $L$ is bounded. With continuity of $L$ you can solve the equation with sequential approximations. Moreover, you can apply the whole theory developed for continuous functions and, in particular, for continuous linear operators. Since continuity is very natural condition when solving differential equations, we require $L$ to be bounded.

As for the coercivity, note that it, in particular, implies injectivity. Injectivity guarantees us uniqueness of the solution $u$ of the equation $Lu=f$. But when you are solving such an equation, it is desirable that solution depends on right hand side $f$ continuously. Well, this property depends on $L$, and it is sufficient to require coercivity of $L$. Speaking functional-analytically bounded coercive operator perform linear homeomorphism between domain and its range. Hence there is a "nice" correspondence between initial data $f$ and solution $u$.

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can you please elaborate why continuity is important because this is the property of an operator, not a solution. – Medan Mar 23 '12 at 23:32

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.

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