368 reputation
110
bio website poisson.phc.unipi.it/~lucia
location Madrid, Spain
age 26
visits member for 3 years, 4 months
seen 10 hours ago

Math PhD student at Universidad Complutense de Madrid, working on Quantum Information and Quantum Dissipative Systems.


2d
comment Show $ \{ (\xi,\eta,\zeta) \in \mathbb{R^3} : \xi = \eta = \zeta \}$ is closed
I never use triangle inequality: I am using the fact that $b_1 - a_1 \le |b_1 - a_1|$ and $b_2 - a_2 \ge - | b_2 - a_2|$. Shall I make it more clear in the answer?
2d
comment Prove $\mathbb{R}^k$ is complete w.r.t. the maximum norm
Try to write down the condition for a convergent/Cauchy sequence w.r.t. the $\Vert \cdot \Vert_\infty$ norm, and see if you can apply your knowledge on completeness of $\mathbb R$...
Dec
16
comment Basis of a vector space is a maximal linearly-independent set?
What is your definition of basis of a vector space? (There is a number of equivalent ways of defining a basis...)
Apr
4
comment Expanding information capacity of Gaussian Channel
Could you please clarify your notation? I suppose $I(\cdot; \cdot)$ is the mutual information, and $h(\cdot)$ is the Shannon entropy, right? Also, what are $X$, $Y$ and $Z$?
Apr
4
comment Weak star limit
Sorry, I do not understand your question: what is $k$ in your definition? You defined $A^\epsilon$, what is $A_n$? (Is this homework?)
Apr
3
comment Exercise Functional Analysis
Also, you should really have a look at en.wikipedia.org/wiki/Derivation_(abstract_algebra)
Apr
3
comment Exercise Functional Analysis
It seems to me that if you define an arbitrary value for $\mathcal O(x)$, then you can extend the operator uniquely to polynomials (and then by density to differentiable functions).
Oct
20
comment Positivity of the anti-commutator of two positive operators implies commutativity?
Yes, indeed. The reason I was expecting them to commute is that if we call $P_B$ the ortogonal projector onto the kernel of $B$, then the fact that $\{ A, B \} \ge 0$ do imply that $A$ commutes with $P_B$. (in your example, this is trivial, but is not in general)
Oct
19
comment Positivity of the anti-commutator of two positive operators implies commutativity?
Nice! I was pretty convinced that it was true.
Oct
19
comment Positivity of the anti-commutator of two positive operators implies commutativity?
In your examples $A$ and $B$ are not Hermitian, and thus not positive, at least under one definition of positivity. Which is probably the most common. Where you thinking of the generalization of positivity for non hermitian matrices?
Aug
30
comment Functional from $(0,+\infty)$ to $L^p(\mathbb R)$
It's ok: it was probably the only possible interpretation, and in fact looks like nobody misunderstood what you meant.
Aug
29
comment Functional from $(0,+\infty)$ to $L^p(\mathbb R)$
I am trying to understand how $f^r$ is defined: maybe you wanted to say that for every $r \in (0, \infty)$, $f^r$ is in $L^p(\mathbb{R})$. That's different from saying that for every $r$ we have $f^r : (0,\infty) \to L^p(\mathbb{R})$.
Aug
24
comment Positive functionals on $\ell^\infty$
I had looked for questions about "positive operator", but I forgot to look for "non-negative operators"... anyway, I'm perfectly fine with using AC, and I fact I edited my question adding the Banach limits as another class of positive operators. Thank you for the links!
Aug
22
comment Is $\ell^1$ isomorphic to $L^1[0,1]$?
@Theo Buehler : your comment indeed seemed off-topic to me, but since it was a "nitpick", I didn't dare to question it... maybe you could edit the question title, setting it to something more meaningful and clear of "Basic Functional Analysis Question". Probably "Is $\ell^1$ isomorphic to $L^1$" would fit, or something similar...
Aug
19
comment A locally injective but non globally injective function?
$U$ was supposed to be open, but as your example shows, the interesting question arises when $U$ is connected.
Aug
19
comment A locally injective but non globally injective function?
@Didier Piau: you are right, I was thinking about continuos functions.