Angelo Lucia
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 Jan 6 comment Maximize $\text{trace}(Z^{T}A Z)/\text{trace}(Z^{T}B Z)$ Are they real or complex? What size do they have? What constrains do you have on the maximum? Your question still does not make sense to me. Jan 6 comment Maximize $\text{trace}(Z^{T}A Z)/\text{trace}(Z^{T}B Z)$ What are A,B,Z? What do you mean by $Z^\prime$? Over what set are you considering the maximum? You should clarify your question by explaining your notation. Dec 17 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? 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.