New answers tagged

1

Your argument is correct. Here is a another possibility to find those $g_i$'s. Let $c\ge0$. Define $u_c$ as the weak solution of $$ -\Delta u + cu = F $$ in $H^1_0(\Omega)$. Then $g_0=cu_c$ and $g_1=\nabla u_c$. Note that it is possible to vary $g_0$. But then $g_1$ has to be adapted. It is not possible to only vary $g_0$ (and let $g_1$ be fixed).


1

This depends on $f$. A basic example is any function in $H^1_0$, which means that the trace vanishes. For these functions the answer is trivially affirmative; $g=0$. However, the range of the trace operator on $H^1(\Omega)$ is exactly the space $H^{1/2}(\partial \Omega)$. On the other hand, the range of the trace operator on $H^2(\Omega)$ is the space $H^{3/...


0

First of all stating that $u = 0$ on $\partial U$ and stating that a solution has to lie in $H_0^1(U)$ is superfluous (you could either remover the first or state $u \in H^1(U)$). It does not really matter, just a small remark. Second, I will assume that $\partial U$ is (piecewise) Lipschitz/smooth enough that we can use integration by parts. Then, the ...


1

The idea is indeed to approximate $u$ by smooth functions, and to let $r\to\infty$. To deal with the integral over $\partial B_r$, you do not need to evaluate the integral - it suffices to show it tends to $0$ as $r\to\infty$. The basic reason is that you can choose an approximating sequence $(u_n)$ to $u$ with each $u_n\in C^\infty\cap H^1$ such that $u_n(x)...


1

No, this inequality cannot hold for any $f\in H^1(\mathbb{R})$. When you wonder if such a functional inequality can hold, a standard test is the scaling argument. Here is how it goes. Let $f\in H^1(\mathbb{R})$ be non-identically $0$ and define $f_\lambda(x) = \sqrt{\lambda}f(\lambda x)$ with $\lambda>0$. Then one easily computes $\|f_\lambda\|_{L^2} = \|...


0

This cannot work for just $C^\infty$-functions. Take $u(x):=\frac1{h^2-\|x\|^2}$. It is infinitely often differentiable and unbounded on $B_h$. So it cannot be extended to $B_2$.


1

Such an inequality can't hold for scaling reasons: Let, for $a>0$, $f_a(x):= a^{-1/2}f(x/a)$. Then $\| f_a\|_{L^2}=\| f\|_{L^2}$, $\| f_a\|_{L^\infty}= a^{-1/2}\| f\|_{L^\infty}$, and $\| f_a\|_{H^1}= \| f_a\|_{L^2} + \| (f_a)'\|_{L^2}= \| f\|_{L^2} + a^{-1}\| f'\|_{L^2}$. Letting $a\to \infty$ we reach a contradiction if such an inequality held. In fact ...


1

Actually, $W^{r,2}(0,1)$ is a space of equivalence classes of function, i.e, they are only defined a.e. and you can simply change the function on a set of measure zero but it identifies the same function in $W^{r,2}(0,1)$. Now, note that $W^{1,1}(a,b)$ is embedded in the space $AC([a,b])$, e.g., see here. That means you can identify the function with its ...


1

Yes, this is possible. Sketch of a proof: This can be done with the help of the accepted answer to this question, but it requires some modifications. Let $\varepsilon>0$. In the method in the linked answer one can also choose a function $\phi:\Bbb R\to\Bbb R$ such that $\phi(x)=x$ for $x\in [-1,1]$ and $|\phi(x)|<1+\varepsilon$ for all $x\in \Bbb R$. ...


3

Yes, that's exactly it. Technically this proof establishes the estimate on a dense subspace of $H^s$ such as Schwartz space, because you are assuming that you can write $u$ as the inverse Fourier transform of an $L^1$ function, whereas the Fourier transform on $L^2$ (which the Fourier characterization of $H^s$ depends on) is defined through a limiting ...


2

Yes. Let $\psi \in C_c^\infty(\Omega)$ with $\int_\Omega \psi \, \mathrm{d}x = 1$ be given. For any $f \in L^2_m(\Omega)$ take a sequence $(f_n) \subset C_c^\infty(\Omega)$ with $f_n \to f$ in $L^2(\Omega)$. Then, $\int f_n \, \mathrm dx \to \int f \, \mathrm dx$. Thus, we can define $$ g_n := f_n - \psi \, \int_\Omega f_n \, \mathrm dx$$ and obtain $g_n \to ...


1

The problem with your argument is that you need the claim to work for $f-g$ in order to proof that it works for $f$. Instead, approximate $f$ by a sequence of smooth functions $(g_n)$. Then by the inequality for these smooth functions, one finds that $ \left( \frac{g_n}{|\cdot|} \right) $ is a Cauchy sequence in $L^2$. Then it remains to prove that the ...


1

A good resource for all things Sobolev-space related is Sobolev Spaces by R. A. Adams and J. J. F. Fournier, $2^\mathrm{nd}$ edition. Page $60$ there provides the definition that you're looking for: $$H^m_q = H^{m,q}(\Omega) \equiv \mbox{the completion of } \{u\in C^m(\Omega) \mid \|u\|_{m,q} \lt \infty \} $$ where the completion is with respect to the ...


1

Since $f(|x|, u)$ is a composition of continuous functions (up to the boundary), then $f$ is continuous up to the boundary, whence $L^\infty(\overline \Omega)$ and therefore $L^p(\overline \Omega)$ for all $p \geq 1$. In particular, $f \in L^2(\Omega)$. Then, by Theorem 8.12 in Gilbarg-Trudinger, $u \in W^{2, 2} \Omega$. By the Agmon-Douglis-Nirenberg ...


0

If $u,v\in H^2(\Omega)\cap H_0^1(\Omega)$, then $$ \int_{\Omega}u\Delta v\,dx=-\int_{\Omega}\nabla u\cdot\nabla v\,dx. $$ Hence $$ \int_{\Omega}u\Delta u\,dx=-\int_{\Omega}|\nabla u|^2\,dx $$ and thus $$ \|u\|_{L^2}\|u\|_{H^2} \ge \int_{\Omega}u(u-\Delta u)\,dx=\int_{\Omega}(u^2+|\nabla u|^2)\,dx=\|u\|_{H^1}^2 $$ which implies tha


1

In the following I'll refer to your first space as X and to the second as Y. As the question is set in Lebesgue-Spaces $L^2$, $u'$ means the weak derivative of $u$ which is defined by $u'\in L^2$ and $$ \int u(t)\phi'(t) dt = - \int u'(t)\phi(t)dt$$ for all $\phi\in C_c^1$. I'll just assume, that you forgot the '-' in your equation. So let $u$ be in X. ...


0

You said you can do b) with compact [a,b]. This approach can be salvaged because you can extend an $H^1$ function from $(a,b)$ to $[a,b]$ as a (Hölder) continuous function. For c) try to construct a radially symmetric singular function whose $H^1$ norm is finite.


0

I don't think that this will be true. Let us ignore the time-dependence and replace $\varepsilon(u_n)$ by $f_n$. Then, you essentially ask whether the weak convergence $f_n \rightharpoonup f$ implies $$A(f_n) \rightharpoonup A(f)$$ for some non-linear, pointwise defined mapping $A$. This is, however, not true. Let us take $$A(f) = \begin{cases} f & \text{...


0

The absolute continuous part of the assertion is somewhat useless, it is just here to say that u' makes sense for people who do not know distribution theory. Nevertheless your question boils down to the fact that $H^{3/2}$ is imbedded into absolutely continuous functions. The proof depends on how you define $H^{3/2}$. If for example you follow the ...


1

You wrote $ \int_0^1 |f|^2 dt + \| f\|$ for $ \int_0^1 |f|^2 dt + \| f\|^{2}$ $f(x)=\int_0^{x} f'(t)dt$ so $|f(x)| \leq \int_0^{1}|f'(t)| dt\leq \sqrt {\int_0^{1}|f'(t)|^{2} dt}$ This implies that $\int_0^{1}|f(t)|^{2} dt \leq \int_0^{1}|f'(t)|^{2} dt=\|f\|^{2}$. Here is a different approach to the first part. Define $T(f)=f(\frac 1 2)=\int_0^{\frac 1 2} f'(...


1

You can use a subsequence-subsequence argument: Assume that $u_n$ does not converge towards $u$ in $C(\bar I)$. Then, there is a subsequence $u_{n_k}$ with $\| u_{n_k} \ge u\|_C \ge \varepsilon > 0$. But your argument shows that this subsequence has a subsequence which converges towards $u$ in $C(\bar I)$. This is a contradiction.


4

Let $f \in \mathcal{S} (\mathbb{R})$. Let $x^*$ be any point realizing the maximum of $|f|$. Then $$2|f(x^*)| = \left|\int_{-\infty}^{x^*} f'(t) \ dt\right| + \left|\int_{x^*}^{+\infty} f'(t) \ dt\right| \leq \|f'\|_{\mathbb{L}^1}.$$ Hence, $\|f\|_{\mathbb{L}^\infty} \leq \frac{1}{2} \|f'\|_{\mathbb{L}^1}$. Applying this inequality to $f^2$ and using the ...


0

Your blf as in the comment, i.e., $$a(u,v) = \int_\Omega \vartheta \nabla u \cdot \nabla v -\int_\Omega (bu) \cdot \nabla v $$ for all $v \in H_0^1(\Omega)$ is correct. It is bounded in $H_0^1 \times H_0^1$ since $$\begin{aligned}|a(u,v)| &\leq \vartheta \|\nabla u\|_{L^2} \|\nabla v\|_{L^2} + \|b\|_{L^\infty} \|u\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\...


1

Note that $\int u \phi_j = - \langle u_j, \phi\rangle_{D' \times D}$ where $u_j \in D'$ is the distributional derivative of $u$. You know $|\langle u_j, \phi \rangle_{D' \times D}| \leq C|\phi|_{L^q}$ for all $\phi \in D \subset L^q$, which means $\sup_{\phi \in L^q} \frac{|\langle u_j, \phi \rangle_{D' \times D}|}{|\phi|_{L^q}} \leq C$. Now, note that this ...


0

I know this is an old question, but I think it's worth to improve the previous answer with some remarks for the benefit of future readers. The space $\{ \mathbf u \in \mathbf{H}(\operatorname{\textbf{curl}}; \Omega) \cap H(\operatorname{\textbf{div}};\Omega) : \mathbf u \cdot \mathbf n |_{\Gamma_1} = 0, \ \mathbf u \times \mathbf n |_{\Gamma_2} = \mathbf 0 \...


1

One criteria follows from the following. Theorem: If for $u \in H^k([0,\infty))$ with $k \leq 2,$ if we define $$ Eu(x) = \begin{cases} u(x) & \text{ if } x \geq 0, \\ u(-x)+3u(-x/3)-3u(-2x/3) & \text{ if } x < 0, \end{cases}$$ then $Eu \in H^k(\Bbb R),$ with comparable norms. It may be possible to use a simpler extension here, but I ...


1

$\newcommand{\R}{\mathbb{R}}$ Fix $\xi \in \R^n$ of length one and denote $$ w_h(y) := \frac{w(y+h\xi)-w(y)}{h} \quad \text{for } h \in \R. $$ If the rest of the argument is fine, we have $w_h \to \frac{\partial w}{\partial \xi}$ in $L_{\operatorname{loc}}^1(\Omega)$ as $h \to 0$. Focus first on $h > 0$. Since $w = 0$ on $\Gamma_u$ and $w \ge 0$ in ...


1

If $u_n\rightarrow u$ in $W^{1,p}_0,$ then (in particular) it converges in $L^p$. If $u_n\rightarrow u$ in $L^p$, then there exists a subsequence $(u_{n_j})$ of $(u_n)$ with the property that $u_{n_j}\rightarrow u$ pointwise almost everywhere. The proof of the latter fact comes in when demonstrating that $L^p$ is complete.


0

My assert follow immediately by the convexity of the function $x\mapsto x^2$.


0

Let me reach a bit further back. Your notation is a bit sloppy, which may mislead and thus confuse you. The correct expression for the differential operator is $\nabla \cdot (k \nabla u) = \operatorname{div} (k \nabla u)$, i.e. the divergence of $k \nabla u$, while $\nabla (k \nabla u)$ would denote the gradient of the vector field $k \nabla u$, i.e. the ...


0

I believe you would need $k \nabla u \in H^1$ to have these defined directly. Assuming that $k \in L^\infty$ and $\nabla u \in H^1$ in general isn't enough to make the product in $H^1$ (take $k$ to be very not-differentiable, but still bounded, and take $\nabla u$ to be nonzero where $k$ is badly behaved). However, you could assume the weaker condition that $...


1

In the computation of $\int_{\Omega}|\nabla f|^2 \,dx$ you forgot $r\,dr$. Hence the integral converges since $\frac{1}{\log(1+1/r)^2 r(1+r)^2}$ is integrable close to $r=0$ and we get $f\in H^1(\Omega)$.


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