Daniel Spector
• Member for 9 years, 2 months
• Last seen more than 7 years ago
• Haifa, Israel

I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An ...

I can think of two differing viewpoints, and though many would prefer the latter due to its simplicity inherited from the Hilbert structure, I prefer the former for its connections to Sobolev spaces ...

Indeed, the norm is one. To see this, fix a cutoff function $\phi \in C^\infty_c(U)$ (which we only need if $U$ is unbounded, to make sure the constructed functions are integrable) and define $... View answer 4 votes The answer is no - take the sequence$f_n(x): n^\frac{1}{3} \chi_{[0,\frac{1}{n}}(x)$. Then$f_n \to 0$in$L^2$strongly but$TV(f_n) = 2n^\frac{1}{3} \to +\infty$. In general, an arbitrary$f \in ...

The reason for a Lipschitz condition is to guarantee uniqueness, as a standard example of non-uniqueness when $f$ is not Lipschitz is $f(x,y)=y^{1/3}$ with the initial condition $y(0)=0$. Then $y=... View answer Accepted answer 3 votes An equivalent formulation of your question is to prove that$f:X^\prime \to \mathbb{R}$be written as the supremum of affine functions, i.e.$f(x^\prime)=\sup_{i \in \mathbb{N}} \langle x^\prime,x_i\...

Suppose $f \in W^{1,1}_{loc}(U)$. Then no, since for such an $f$, we have that $Df$ exists and the approximate limit $ap\lim_{y\to x} \frac{f(x)-f(y)-Df(x)(x-y)}{|x-y|} = 0$ exists for almost every ...

I believe the following is a counterexample: Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$, and $u^\prime(x) = 2H(x)-1$, where $H$ is the ...

Here is what seems to me a good beginning, if the argument can be carried all the way. We have that $dist(\mu_1,\mu_2):=\inf_c \sup_s f(s,c)$, where $f(s,c):= c( \{ (\omega^1,\omega^2) \in X\times ... View answer 2 votes I think it would help to specify what you mean by subregion$\omega \subset \Omega$. In general, it is absolutely true for$\omega$a lower dimensional object. For example, when$n=2$and$u \in H^1(...

The result you want is true because of existence and Uniqueness for second order non-linear ODE, for example see Boyce and Diprima. Writing $y^{\prime\prime}=f(t,y,y^\prime)$ and verifying that $f$, ...

There may be a name for this, but it seems like a strange condition. Such a function cannot take a constant value on any set of positive Lebesgue measure, otherwise the inverse image of that constant ...

In general, whenever you have a (separately) convex function $f :\mathbb{R}^N \to \mathbb{R}$, which means it is convex in each variable, this implies the function is locally Lipschitz. The fact that ...

The only smooth solution on some smooth, open domain in $\mathbb{R}^2$ is the zero solution. Consider the following: If $f$ solving this equation were smooth, then we would require $f_{xy}=f_{yx}$,...

I think the first thing to iron out is that $\Omega$ should be open, since compact would be closed and then even defining $H^1(\Omega)$ is not trivial. Maybe assume $\Omega \subset \mathbb{R}^N$ is ...
Perhaps if you consider the case where $\Omega_j$ are sets of finite perimeter you can recover the statement. A set $\Omega_j$ is of finite perimeter in $\Omega \subset \mathbb{R}^N$ if $\chi_{\... View answer 1 votes Perhaps you have simplified this from a multidimensional radial problem, which explains the constraint$\int_0^B u^k(t) t^{h-1}\;dt = C$, which if we view$u$as a radial function in$h$dimensions is ... View answer 0 votes Ok. So I thought a little more carefully about it, though I still have not seen the Lions-Magenes book, but I think$C^\infty$boundary is related to the above result holding for more general spaces (... View answer Accepted answer 0 votes If you assume you have proven the Radon-Nikodym theorem for measures$\mu,\nu$,$\nu <<\mu$, then for any$F$in such a class, the measure$\mu_F,\nu_F$defined by$\mu_F(E):=\int_E F(x)\;d\mu$... View answer 0 votes Ok. Suppose$\mathcal{F}_n \subset L^2(\Omega)$such that$\mathcal{F}_n$is spanned by$n$linearly independent "vectors" with bounded norm, which means$L^2(\Omega)$functions such that$||f||\leq ...