bio | website | professor.ufabc.edu.br/… |
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location | Cotia, Brazil | |
age | 36 | |
visits | member for | 4 years, 7 months |
seen | 7 hours ago | |
stats | profile views | 1,011 |
I'm a professor at the Center of Mathematics, Computation and Cognition of the Federal University of the ABC, Santo André, Brazil.
Jun 17 |
revised |
Do locally convex topological vector spaces embed into diffeological spaces?
Added reference to complementary answer |
Jun 17 |
comment |
Exponential rule for Whitney-$\mathcal{C}^{\infty}$-topology
@JochenWengenroth or perhaps a "convenient-calculus" tag... |
Jun 17 |
comment |
Exponential rule for Whitney-$\mathcal{C}^{\infty}$-topology
@KathrinL. Igor's comment is still relevant. The graph topology on the space of continuous maps is the Whitney $C^0$ topology, and Peter Michor's argument works in the continuous case as well. |
Jun 16 |
comment |
Do locally convex topological vector spaces embed into diffeological spaces?
I have edited my answer to better suit your original question, based on your comments. I've also added a remark at the end concerning your remaining question, to which unfortunately I don't know an answer. Papers who cite Glöckner's work quoted above, according to MathSciNet, apparently don't even raise that question, let alone answer it. |
Jun 16 |
revised |
Do locally convex topological vector spaces embed into diffeological spaces?
Improved explanation in view of OP's comments, superfluous discussion on continuous linear maps removed |
Jun 15 |
comment |
Do locally convex topological vector spaces embed into diffeological spaces?
So, you were considering smooth maps (in either sense) as the arrows of the category of lctvs in your original question, and your residual question is whether the answer changes if you restrict the arrows to be smooth diffeomorphisms. Is this corrrect? |
Jun 15 |
revised |
Do locally convex topological vector spaces embed into diffeological spaces?
Improved explanation in view of comments by the OP |
Jun 15 |
revised |
Do locally convex topological vector spaces embed into diffeological spaces?
small rewording |
Jun 15 |
comment |
Do locally convex topological vector spaces embed into diffeological spaces?
@DavidRoberts so, what is the definition of "embedding of categories" you have in mind in your question? Do you put any restriction on the arrows in the category of lctvs besides being continuous linear maps (e.g. do you require them to be topological isomorphisms, or something)? |
Jun 15 |
revised |
Do locally convex topological vector spaces embed into diffeological spaces?
Conclusion rectified in order to match previously incompletely understood hypothesis, reasoning unchanged |
Jun 15 |
answered | Do locally convex topological vector spaces embed into diffeological spaces? |
Jun 6 |
comment |
constant rank theorem for banach spaces
Interesting. This means that the hypothesis that $M$ is finite-dimensional also removes the requirement that $\ker DF[u_0]$ should be a closed direct summand. That is very useful. Curiously, Glöckner's paper does not quote the book of Abraham et al. |
Jun 1 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
Notice, however, that you can use Stokes's theorem only if the boundary of the region is at least locally Lipschitz. |
Jun 1 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
I apologize, actually smooth functions / forms of compact support are dense only in $L^2$ for the region you specified. For $H^k$ you need to specify another dense domain - a possibility would be to choose smooth functions / forms in $H^k$ whose derivatives up to order $k$ extend continuously to the boundary, and then keep track of the boundary terms that appear due to Stokes's theorem. In this case, $\Box_g$ will no longer be symmetric in general due to those. Of course, this is only one choice of domain - the problem you are studying may impose another choice. |
Jun 1 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
It applies to both, since the domain of $\Box_g$ is assumed to be the (dense in both) linear subspace of smooth functions / forms with compact support. Again, $\Box_g$ is not defined as a linear operator from either Hilbert space into itself. |
May 31 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
By the way, what is your definition of $\mathcal{U}^+_t$? |
May 31 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
In other words, the formal adjoint of $\Box_g$ (i.e. the restriction of the adjoint of $\Box_g$ to the latter's domain) is equal to $\Box_g$ itself. |
May 31 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
Computing the adjoint of an unbounded operator such as $\Box_g$ involves not only finding its formula, but also its domain from that of $\Box_g$. It's not clear from your reasoning which is the domain you assume for $\Box_g$ - notice that it cannot be the whole Hilbert space, be it $L^2$ of $H^k$. If you assume, as in my former comment, its domain to be smooth forms of compact support, then the symmetry of $\Box_g$ follows immediately from the definition, since $\Box_g=d\delta+\delta d$, where the co-differential $\delta$ is the formal adjoint of $d$. |
May 31 |
comment |
How can I calculate the adjoint of the wave operator $\square_{g}$ in $H^{k}$?
Just a few remarks: the Laplace-Beltrami operator $\Box_g$ on a Riemannian manifold $(M,g)$ is essentially self-adjoint with domain of smooth $p$-forms with compact support when $(M,g)$ is complete (particularly when $M$ is compact), as shown by Chernoff, Strichartz and other people. In complete generality, $\Box_g$ has at least one self-adjoint extension - namely, the Friedrichs extension, since the quadratic form associated to $\Box_g$ is bounded from below. The latter property is lost when $g$ is not positive definite. All we can say then is that $\Box_g$ is symmetric. |
May 25 |
comment |
John Nash's Mathematical Legacy
@DennisSerre Newton iteration was originally designed for finding simple zeros of functions - indeed, in this case a simple zero of (say) a real valued function $f$ on the real line will be a fixed point of the map $g(x)=x-f(x)/f'(x)$. In Nash-Moser's method, $g$ is replaced at the $n$-th step by a parameter-dependent regularization $g_{\theta_n}$ of $g$, where $\theta_1<\theta_2<\cdots\rightarrow+\infty$ is an increasing sequence of scales, so it's not really a fixed-point method. |