# Show this bilinear form is coercive

Suppose $a(u,v):V\times V$ is a bounded, symmetric and coercive bilinear form. $V$ is eg. $H^k(\Omega).$

Let $g \in C^\infty(\Omega)$ be such that $A \leq g^{(k)} \leq B$ for positive $A$, $B$ and $k$.

Is there any way to show that the bilinear form $b(u,v) := a(u, vg)$ is coercive? i.e., that $$a(v,vg) \geq C\lVert v \rVert^2?$$

I tried a lot of things for my particular problem but they all give me coercivity IF certain constants satisfy some condition. Perhaps there is a way to do it abstractly without going into the details?

I ask this because I have not yet solved Existence of solution for this parabolic PDE in the affirmative.

• Probably not. The bilinear form $b$ will in general not only depend on $g$, but also on it's derivatives up to order $k$. A bound on the sup-norm doesn't give you any information on $g^{(k)}$ for $k>0$ (or at least not very much). – Sam Dec 28 '12 at 11:29
• @SamL. Suppose $g$'s derivatives are also bounded similarly. – soup Dec 28 '12 at 11:31
• The condition on derivatives is indeed necessary, otherwise something like $a(u,v)=\int_0^\pi (2u'v'-uv)$ on $H_0^1(0,\pi)$ would yield a counterexample. Of course, now that we have to talk about derivatives, we can't solve the problem in a totally abstract way. The bound on derivatives potentially helps if $a$ is associated with a local differential operator, e.g., $a(u,v)=\langle Du,v\rangle_{H^k}$ where $D$ is a densely defined self-adjoint differential operator. Does $a$ have such a structure? – user53153 Dec 28 '12 at 17:15
• @PavelM Thanks for the attention. In fact $a(u,v) = (Du, Dv)$ where $D$ is 2nd order elliptic,say the Laplacian. Sadly I can't see any way to get this result. I have to use the chain rule to rewrite $D(vg)$ which introduces a gradient term which messes things up. – soup Dec 28 '12 at 17:23
• But those additional terms will be of lower degree (fewer derivatives on $v$ than in the form itself), so you should be able to control them. For example, the $L^2$ norm of $\Delta u$ controls $L^2$ norms of all second derivatives (Evans 1st ed., section 6.3.1). The second derivatives bound lower orders, via Poincare's inequality. – user53153 Dec 28 '12 at 17:30