Let $u\in\mathcal{C}^2(\Omega)\cap\mathcal{C}^0(\overline{\Omega})$ be a classical solution of $\Delta u=u^3-u$ on a bounded domain $\Omega\subset\mathbb{R}^2$. Assume that $u=0$ on $\partial\Omega$. Then it holds that $|u|\leq 1$ on $\Omega$. Does someone know how to prove that?
59194
Let us consider that $u\in C^2(\Omega)\cap C^0(\Omega)$ and satisfying the following equation .  $\Delta u=u^3-u , x\in\Omega$ and   $u=0 $   in  $\partial \Omega$  $\Omega \subset \mathbb R^n$ and bounded .   I need hints to find out what possible value $u$ can take ?  Thank you for ur hints in advance .
75175
Let $\Omega$ be a bounded connected open subset of $\mathbb{C}$ containing $0$. Let $f: \Omega \rightarrow \Omega$ be holomorphic and $f(0) = 0$, $f'(0) = 1$. The problem I am working on is to show that $f(z) = z$.  If $\Omega = \mathbb{D}$, then this follows from the Schwarz Lemma. I also know of a solution () which involves looking at the power series coefficients of $f^n := f\circ f \circ f \circ \cdots \circ f$ ($n$ times) and using the Cauchy estimates, but is there a different way of doing this problem that doesn't involve taking the $k$th derivative of $f^n$?
8509
Let $(M,\omega)$ be a Kähler manifold. Why is the Kähler condition $$d \omega = 0$$ equivalent to $$\partial_kg_{i\bar j} = \partial_ ig_{k\bar j}$$ for all $i, j, k$?  I am looking for a reference.
98683
If I have a manifold $M$ and $\omega \in \Omega^q(M)$, considered an open subset $U$ of $M$, and choose a point $p \in M$, is it true that $(\omega|_{U})_p=0 \rightarrow \omega_p=0$? With $\omega|_{U}$ I mean the pull-back of $\omega$ by the embedding $F:U\to M$. The converse is obviously true, but this implication?
90153
The following is an excerpt from Folland's Introduction to Partial Differential Equations:  Let the open set $\Omega\subset{\mathbb R}^n$, and $k$ be a positive integer. $C^k(\Omega)$ will denote the space of functions possessing continuous derivatives up to order $k$ on $\Omega$, and $C^k(\overline{\Omega})$ will denote the space of all $u\in C^k(\Omega)$ such that $\partial^{\alpha}u$ extends continuously to the closure $\overline{\Omega}$ for $0\leq|\alpha|\leq k$.  As I understand, an extension means that there exists $\widetilde{\partial^{\alpha}u}$ which is defined on $\overline{\Omega}$ and $\widetilde{\partial^{\alpha}u}|_{\Omega}=\partial^{\alpha}u$. And "extends continuously" means $\widetilde{\partial^{\alpha}u}$ is continuous with respect to the relative topology on $\overline{\Omega}$.  Here are my questions:        How do people usually do such extension?   When does such extension exist and when does not?   Let $\Omega$ be an open subset of ${\mathbb R}$. Is there any   non-trivial counterexample such that   this kind of extension does not exist?
38819
Let $\Omega$ be a region in the complex plane and let $f_1$ and $f_2$ be holomorphic functions on $\Omega$ having no common zero. Show that there exist holomorphic functions $g_1$ and $g_2$ on $\Omega$ such that, $f_1g_1+f_2g_2$ is identically equal to one on $\Omega$.
49098
There is  with this exact same prompt which got several down-votes for not showing  their work.  So I'll show what work I've got.  I know being a harmonic function implies satisfying the Mean Value Property, thus what I thought I'd do is consider two arbitrary points in $\mathbb{C}$ and prove that:  $$|u(z) - u(w)|=|\frac{1}{2\pi}\int_0^{2\pi}u(z - re^{i\theta})d\theta - \frac{1}{2\pi}\int_0^{2\pi}u(w - te^{i\theta})d\theta| $$ $$= |\frac{1}{2\pi}\int_0^{2\pi}u(z - re^{i\theta}) - u(w - te^{i\theta})d\theta|$$  Now what I wan't to do is take r and t to infinity and prove this integral goes to zero, but I can't figure out how to do that.  I've used the fact that my harmonic function u is defined on all of $\mathbb{C}$ in taking t and r to infinity but I still need to use the fact that it's bounded.  I also read through the proof of Liouville's Theorem, given as an answer in that other post, which seems intuitively correct but rather un-rigorous.  It actually says that the value in the center of a ball is equal to the average over the ball's volume.  I imagine this can be proven from the average over the boundary since the average over the boundary will not change as you continuously shrink the boundary down to its center. This however would require the difference of two surface integrals, or is there a simpler way?   I was also considering using the Maximum Modulus Principle somehow but I haven't found a way yet.  I've probably made this way more complicated than it has to be, hopefully someone can help me, thanks! =].  Edit: Ok after thinking some more about the accepted answer in that other post, I've decided what I need to do is look at the canonical holomorphic function whose real part is my harmonic function u, and prove that holomorphic function is bounded.  From there I can prove that since it's entire and bounded that it's constant, and from there prove that then its real part must be constant.  So could someone talk a bit about the construction of this canonical holomorphic function?
46921
Show that the Cauchy problem $x''=x|x|,x(0)=a,x^{\prime}(0)=b$,has a unique solution for all $a,b\in \Bbb R$. (The Problem is from A text book on ordinary differential equation by Shair Ahmad and Antonio Ambrosetti,exercise 4.3,pg:70)  My attempt:  I am using this theorem(given in book,page 69):  $\begin{cases} x^{(n)}=f(t,x',x'',...,x^{(n-1)})\\              x(t_0)=\alpha_1,x'(t_0)=\alpha_2,...,x^{(n-1)}(t_0)=\alpha_n      \end{cases}$                                         (1)  If $f:\Omega \to \Bbb R$ is continuous and locally lipschitzian with respect to $(x_1,x_2,...,x_n),$ then (1) has a unique solution,defined in a suitable nbd of $t_0$.($\Omega$ is an interval containing point $(t_0,\alpha_1,...,\alpha_n)$)  (The function $f$ is called locally lipschitzian at  $(t_0,\alpha_1,...,\alpha_n) \in \Omega$ with respect to $(x_1,x_2,...,x_n)$ if $\exists$ a neighborhood $U \subset \Omega$ of $(t_0,\alpha_1,...,\alpha_n)$ and a number $L&gt;0$ such that, denoting by $|.|$ the euclidean norm in $\Bbb R^n$, one has   $|f(t,x_1,x_2,...,x_n)-f(t,z_1,z_2,...,z_n)|\leq L |(x_1,x_2,...,x_n)-(z_1,z_2,...,z_n)|$ for every$(x_1,x_2,...,x_n),(z_1,z_2,...,z_n) \in U$. )  clearly $f$ is continuous.therefore we will show that $f$is lipschitzian at point $(0,a,b)$.  To prove lipschitzian at point $(0,a,b)$ we have to show that $\exists$ a nbd U of $\Bbb R^3$ containing point$(0,a,b)$ and a number $L&gt;0$ such that  $|f(t,x,x_1)-f(t,z,z_2)| \leq L||(x,x_1)-(z,z_1)||$   $\forall (t,x,x_1),(t,z,z_1) \in U$  here $||(x,x_1)-(z,z_1)||=\sqrt{(x-z)^2+(x_1-z_1)^2}$  Now $x|x|$ is differentiable everywhere on $\Bbb R$.therefore it is locally lipschitz on any finite interval of $\Bbb R$. therefore we have (for finite interval)  $|x|x|-z|z||\leq L|x-z|$ for some number $L$.  Now   $|x|x|-z|z||\leq L|x-z| \leq L \sqrt{(x-z)^2+(x_1-z_1)^2}$  $\therefore$$|f(t,x,x_1)-f(t,z,z_2)| \leq L||(x,x_1)-(z,z_1)||$   $\forall (t,x,x_1),(t,z,z_1) \in U$ where $U$ is any nbd of finite radius of $\Bbb R^3$ which contains the point $(0,a,b)$  Hence given problem has a unique solution for all $a,b\in \Bbb R$.   I am confusing whether am I correct or not?If it is not correct then where am I making mistake?and how to solve this problem?Thank you in advance.
108752
I am working on the following problem from Stein and Shakarchi:  Let $f$ be an integral function on $\mathbb{R}^d$ such that $\|f\|_{L^1} = 1$ and let $f^*$ by the Hardy-Littlewood maximal function corresponding of $f$.     Prove that if $f$ is integrable on $\mathbb{R}^d$, and $f$ is not identically zero, then $f^*(x) \geq c/|x|^d $ for some $c &gt; 0$ and all $|x| &gt; 1$.   Conclude that $f^*$ is not integrable on $\mathbb{R}^d$.   Then, show that the weak type estimate $m(\{x:f^*(x) &gt; \alpha \}) \geq c/\alpha$ for all $\alpha &gt; 0$ whenever $\int|f| = 1$, is best possible in the following sense: if $f$ is supported in the unit ball with $\int|f| = 1$, then $m(\{x:f^*(x) &gt; \alpha \}) \geq c'/\alpha$ for some $c' &gt; 0$ and all sufficiently small $\alpha$.     [Hint: For the first part, use the fact that $\int_B|f| &gt; 0$ for some ball $B$.]  I am not really sure where to begin with this.  Any help would be greatly appreciated.
36578
I'm being asked to prove that $$1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$$ where $\omega \ne 1$ is an n-th root of unity, and I don't know where to start  I feel like there's something terribly obvious that I'm missing here.
115312
I have been told by my teacher that, if $f$ is Sobolev (i.e. $W^{1,p}$ for some $p$) on an open set $\Omega$, which, for the proof what I'm asking is needed for, may well be assumed bounded, then its gradient is a.e. zero on each of its level sets, i.e. $\nabla f|_{\{f=k\}}=0$ a.e. for all $k$. I am trying to prove it.  For the moment, I have thought that, calling $A_k=\{f=k\}$, if $A_k$ contains any ball, then the gradient is zero there. This is because I know that, for a.e. line in that ball, $f$ will be absolutely continuous on that line, and since it is constant, its derivative will have to be a.e. zero on a.e. segment, and letting the segment length go to zero I can finally say the derivative at the endpoint is zero.  But what about non-interior points? I don't think I can say much about the gradient there. So I would have to prove that, if I take $A_k$ and remove all the balls it contains, I am left with a zero measure set.  So how do I go about this? Or maybe I am missing something and I can, in fact, say something about the gradient in at least some of those left-out points?  "Edit"  Naturally the ball discussion is overkill: $f$ is constant on that ball, so it is smooth therein, hence weak derivatives are classical, and we know classical derivatives of constants are zero.
90763
Let $z_0$ be a point in an open set $U \subset \mathbb{C}$.  Let $f,g:U \to \Delta(0,1)$ where $g$ is a bijection such that $g(z_0)=0$ and $g'(z_0)\neq 0$. Suppose that $f(z_0)=0$, then $|f'(z_0)|\leq |g'(z_0)|$  I am not even sure how to start this.  I was assuming Schwarz lemma.
29266
In Lieb and Loss's Analysis, I saw that they mentioned  $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $ L^2(\mathbb{R}^n)$ (dense wrt the $L^2$ norm, I think). But I didn't find its proof in the book. So I wonder why it is?    Does this conclusion hold for $L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)$ for any measure space $(\Omega, \mathcal{F}, \mu)$?  Are there similar statements if replace $L^1$ with $L^p$, and  $ L^2$ with $L^q$ for $p \leq q \in (0, \infty ]$ or $\in [1, \infty]$?  Thanks and regards!
44851
Suppose $f$ is a continuous complex valued function on a domain $\Omega$. Suppose $f^2$ and $f^3$ are holomorphic in $\Omega$. Show that $f$ is also holomorphic in $\Omega$.   Assume $f=u+iv$. I see that if $u,v$ are in $C^1$ then $f^2$ is holomorphic can imply $f$ is holomorphic considering Cauchy-Riemann equations. But I don't know how to get $u,v$ are in $C^1$ by the adding condition $f^3$ is holomorphic. (I can't get $u,v$ from some combinations of real and imaginary part of $f^2$ and $f^3$). Do someone know how to do this?
36811
Suppose $K(x,y)$ is a symmetric kernel. Let $\phi\in L^2(\Omega)$, where $\Omega$ everywhere is a domain in $R^n$.  Can $\int_{\Omega}K(x,y)\,\phi(y)\,dy$ belong to $L^2$?  In other words can an inequality of the type  $$ \left\|\int_{\Omega}K(x,y)\,\phi(y)\,dy\right\|_{2} \leq \|K\|_{L^1(\Omega\times\Omega)} \|\varphi\|_{2} $$ exist?
75505
Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. And let $L^2\left(\Omega\right)$ be the space of equivalence classes of square integrable functions in $\Omega$ given by the equivalence relation $u\sim v \iff u(x)=v(x)\, \text{a.e.}$ being a.e. almost everywhere, in other words, two functions belong to the same equivalence classes if they only are different in a zero meassure set.  This space is a Hilbert space with the norm is:  $$ \Vert v \Vert ^2_{L^2 \left( \Omega \right) }= \int_\Omega v(x)^2\,\mathrm{d}x $$  Let $H^1\left(\Omega\right)=\left\{ v\in L^2\left( \Omega \right);\,\vert\mathbf{grad}(v)\vert \in L^2\left( \Omega \right ) \right\}$ be also a Hilbert space and its norm :  $$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) }= \Vert v \Vert ^2_{L^2 \left( \Omega \right) } + \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } $$   I've been asked to prove that:  $$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) } \le C\left( \Omega\right) \left( \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } + \int_\Gamma \left ( \gamma v(x) \right) ^2\,\mathrm{d}\sigma \right) \quad \forall v\in H^1 \left( \Omega \right) $$  where $\Gamma = \partial \Omega $, $\Omega $ is smooth enough, $\mathrm{d}\sigma$ is a meassure on the boundary of $\Omega$, and the operator $\gamma: L^2\left( \Omega\right )\mapsto L^2\left( \Gamma\right )$ is the trace operator. ( if it is needed I can define it also )  Sincerely, I just don't know how to begin, and maybe someone could give me some clue about it.(by the way the $C$ is a positive constant depending only on the domain $\Omega$).  Thank you very much!
67319
Prove that if $\Delta u=0$ in $\Omega\subset \mathbb{R}^n$ and $u=\partial u/\partial\nu=0$ on an open smooth portion of $\partial \Omega$, then $u$ is identically zero.  I found a proof , but I'm not sure if it's the correct proof. Does anyone have any idea? Any hint would be appreciated.