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4

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. ...


2

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 $$ \int_{B_1} \lvert u\rvert = \int_0^{2\pi}\int_0^1 r\lvert u(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= \int_0^{2\pi}\int_0^1 r\lvert U(r)\rvert\,dr\,d\vartheta=2\pi\int_0^1 r\lvert U(r)\rvert\,dr. $$ ...


2

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.


2

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 ...


2

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, ...


1

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): ...


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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 ...


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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 ...


1

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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