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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.

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    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).2012-12-28
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    @SamL. Suppose $g$'s derivatives are also bounded similarly.2012-12-28
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    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?2012-12-28
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    @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.2012-12-28
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    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.2012-12-28
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    @PavelM Well, I get the term $\int \nabla v \cdot \nabla g \Delta v$, and I can't get anything nice which $\geq$ that; I have to take the negative of the modulus of the integrand and use Young's inequality. This is what leads me to get a constraint on the constants. Maybe you were thinking of $\leq$ in your comment above?2012-12-28
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    $\nabla g$ is uniformly bounded, so we are left with $\int |\nabla v||\Delta v|$. You can do Cauchy-Schwarz here and then bound $\|\nabla v\|_{L^2}$ by Poincare in terms of $\|\Delta v\|_{L^2}$. Or, to get a little more fancy, $\int |\nabla v||\Delta v| \le \epsilon \int |\Delta v|^2 + 4\epsilon^{-1}\int |\nabla v|^2$ so you don't waste as much of the highest order terms in this estimate.2012-12-28
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    @PavelM Forgive me if I'm being stupid but these are upper bounds you write above, for coercivity we need lower bounds. Of course we can use these upper bounds to get a lower bound but then we have negatives everywhere.2012-12-28
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6875/discussion-between-pavel-m-and-soup)2012-12-28

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