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A mollifier is a function $f$ that you convolve with another function $g$ to get a function which is "close" to $g$ but "nicer". For instance $f$ might be a general $L^1$ function and $g*f$ might be a smooth, compactly supported approximation to $f$. Really a mollifier is not one function but a sequence, or even sometimes a one-parameter continuous family. ...

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The functions $U$, $U_x$ and $U_y$ are locally bounded, except possibly at $(x,y)=(0,0)$. If $B_1$ is the unit ball centred at $(0,0)$, We have $$\int_{B_1} \lvert u\rvert = \int_0^{2\pi}\int_0^1 r\lvert u(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= \int_0^{2\pi}\int_0^1 r\lvert U(r)\rvert\,dr\,d\vartheta=2\pi\int_0^1 r\lvert U(r)\rvert\,dr.$$ ...

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You already have $u^m \to u$ weak-* in $L^\infty(\mathbb{R}^+; H_0^1(\Omega))$. Hence, $u^m \to u$ weak-* in $L^\infty(0,T; H_0^1(\Omega))$. And this gives $u^m \to u$ weak in $L^2(0,T; H_0^1(\Omega))$. No need for Aubin-Lions here.

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So we're in dimension $n=3$ according to the problem statement. I agree with you that we can get the inequality without the $1/t$ factor basically by Cauchy-Schwarz. But am I being silly or does the stated inequality not hold by scaling? Suppose the inequality were true. Let $f_{t}=f(\cdot/t)$. Then ...

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Remember that $f \in W^{1,1} (\gamma_1)$ means that $f, f' \in L^1 (\gamma_1)$ where $f'$ is understood in the weak sense. Remember also that $f'$ in the Sobolev sense in the same as $f'$ in the distributional sense, therefore let us compute the latter one. Finally, we shall use that $\gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}$. Note, ...

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No, you can't. Consider $\Omega$ being a unit circle on $\mathbb{R}^2$. Then: \int\limits_{\Omega} \sum_{|\alpha|=2} | D^{\alpha}u | = \int\limits_{x^2 + y^2 < 1} \left( \left| \frac{\partial^2 u}{\partial x^2} \right| + \left| \frac{\partial^2 u}{\partial x \partial y} \right| + \left| \frac{\partial^2 u}{\partial y^2} \right| ... 1 To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that (I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k. Since \int (\Delta u) v = \int u (\Delta v) (on test functions, at least), you may use induction and prove that \int (\Delta^k u) v = \int u (\Delta^k v), so \int (\sum \limits _{k=0} ... 1 Let M=\|u\|_{\infty}. Since G' is continuous, there exists a constant K such that |G'(s)|\leq K for all s\in [-M,M]. Thus |G'\circ u|\leq K; it follows that |G'\circ u|^p is bounded by K^p. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ... 1 Even though Matt has solved the original problem properly, I feel obligated to give a proper solution to the fixed version. This is just for completeness: Step 1: By scaling, we may suppose that t=1. This is a calculation just like the one done by Matt above. Step 2: By translating f if necessary, we may restrict ourselves to the evaluation of |T_1 ... 1 The assumption is implicitely used here: \begin{align*} \int_{\mathbb{R}^n}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y &= \int_{B_\varepsilon(0)}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y \\ &= (-1)^{\lvert \alpha \rvert} \int_{B_\varepsilon(0)}\int_{\Omega'} ... 1 This is an easy corollary from Hahn-Banach: Let X be a normed vector space and V \subset X a subspace. Then V is dense, if and only if\forall x' \in X': \Big( \big( \forall v \in V : x'(v) = 0 \big)\Rightarrow x' = 0\Big).$$Hint for the proof: If V is not dense, you can separate the closure of V from any point not belonging to this ... 1 At the beginning of this section, Evans makes the assumptions$$ a^{ij},b^i,c\in L^\infty(U), f\in L^2(U). $$The following inequalities are crucial: if w\in H^1(U), then$$ \|w\|_L^2(U) \leq \|w\|_{H^1(U)}, \|w_{x_i}\|_{L^2(U)} \leq \|w\|_{H^1(U)}. $$This is pretty much immediate from the definition of the Sobolev norm. You may need to tack on ... 1 Rewrite the inequality that you want to prove as$$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$Set v:=|u|^{\frac{p+m-1}{2}}. The LHS above becomes$$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}with q=\frac{2p}{p+m-1} and \alpha=\frac{p}{p+m-1}. Apply the NGN ... 1 The answer is yes if u has zero-boundary values and \Omega is sufficiently regular. This follows from H^2-regularity of \begin{align}-\Delta u &= f \text{ in } \Omega \\ u &= 0 \text{ on }\partial\Omega\end{align} with f = -\Delta u \in L^2(\Omega). Otherwise it may not hold. 1 Its the distributional divergence, that is for v \in L^2(\Omega) we have v = \operatorname{div}u iff\int_\Omega v \, \varphi \, \mathrm{d}x = -\int_\Omega u \cdot \nabla \varphi \, \mathrm{d}x \quad\forall \varphi \in C_0^\infty(\Omega).$$To be compared with https://en.wikipedia.org/wiki/Integration_by_parts#Higher_dimensions. 1 It's false. For a counterexample consider n = k = p = 1, U = (0,1) and f \equiv 1. It is possible to extend Sobolev functions under certain regularity assumptions on the boundary of the set U, but it is delicate. Any book on Sobolev spaces addresses the issue. 1 You can't just choose the function to be identically zero outside U in general, because this can cause the weak derivative to fail to be L^p because of "bad regularity" at the boundary. For instance when n=1, all W^{k,p} functions are actually continuous, so extending f(x)=1 on (0,1) to be identically zero elsewhere certainly does not give a ... 1 For \varphi \in C^{\infty}_c one defines \langle\partial_{x_1}f,\varphi\rangle = -\int_{\Omega}f\partial_{x_1}\varphi. Extending this definition to H^1_0(\Omega) we obtain a linear functional on H^1_0(\Omega). Continuity, i.e. boundedness, follows from Holder's inequality (together with Poincare's inequality, depending on the norm you define on ... 1 I have found the answer in Kobayashi & Nomizu, volume 1, page 124. If K is a tensor of type (r,s), then one may construct a new tensor \nabla K of type (r, s+1), defined by$$(\nabla K) (X_1, \dots, X_s, Y) = (\nabla _Y K) (X_1, \dots, X_s) and thus define inductively $\nabla ^k K$ as $\nabla (\nabla ^{k-1} K)$. Choosing now $K$ to be $f$, a ...

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As stated, existence is trivial: $u\equiv 0$ is a solution. When a nontrivial solution exists, it is not unique since one can multiply it by a constant. But this happens only for some specific $\lambda$: those that are in the spectrum of the Dirichlet Laplacian. You can't infer this kind of structure from Lax-Milgram. Rather, sine Fourier series should be ...

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