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Fundamenta Mathematicae.

Biblioteka Wirtualna Matematyki.

European Digital Mathematics Library.


Apr
16
comment Vector spaces isomorphic, then dual spaces isomorphic
Ok, good. I take back my comment! : )
Apr
16
comment Vector spaces isomorphic, then dual spaces isomorphic
Daniel, there is no inner product, the spaces in the question are Banach spaces. For a proof see e.g. the link I give in the comments to the question.
Apr
15
comment Vector spaces isomorphic, then dual spaces isomorphic
The proof given in this answer here seems fairly short.
Apr
15
comment Vector spaces isomorphic, then dual spaces isomorphic
Alright : ) If you replace calculus with banach-spaces it might get you much better help because different people will look at your question. Just a thought.
Apr
15
comment Vector spaces isomorphic, then dual spaces isomorphic
Plus one. Any reason for not adding the banach-spaces tag?
Apr
13
comment Commutativity of two endomorphims
palio: So what are the missing conditions?
Apr
13
comment Is there any book about inequality?
I'm not sure but I think the Cauchy Schwarz Masterclass is a book about inequalities.
Mar
27
comment Is is true that $||v+w||^2 = ||v||^2 + 2\langle v,w \rangle + ||w||^2$?
Note that it is better style to use \langle and \rangle instead of < and >.
Mar
12
comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\} $ implies range $\mathbb{R} - \{ b\} $?
@Unwisdom Thank you for your comment. I suppose the problem arises when, after mapping $(-\infty,a)$ to $(-\infty,a)$, one is forced to map $(a,\infty)$ to $[a,\infty)$.
Mar
11
comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\} $ implies range $\mathbb{R} - \{ b\} $?
@Unwisdom Thank you. I had a long day, could you please elaborate a bit? I know that continuous maps map connected sets to connected sets but it's not obvious to me this very second why this implies that they also have to map disconnected sets to disconnected sets.
Mar
11
comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\} $ implies range $\mathbb{R} - \{ b\} $?
Is it also possible if $f$ is required to be continuous in addition to being bijective?
Mar
1
comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic?
Ooh, you are right, I missed lots of copies of $\mathbb Z$. Thank you very much for your patient comments. I do not fully understand your comment somehow but it gave me the idea on how to prove that the image is $\mathbb Z \oplus \mathbb Z_2$: define a map $\mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus\mathbb Z$ that maps $\langle (0,2) \rangle$ to $(0,0)$ and whose image is $\mathbb Z \oplus \mathbb Z_2$. Like e.g. $f((n,k)) = (n, k \mod 2)$. My only problem now is that I don't know how to figure out what the image looks like. How did you figure out it's $\mathbb Z \oplus \mathbb Z_2$?
Feb
28
comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic?
Just a quick question to see if I am "seeing" the groups in this question correctly: The group $\mathbb Z \oplus \mathbb Z$ contains $3$ copies of $\mathbb Z$ -- one looks like $(k,0)$, one like $(0,k)$ and one like $(k,k)$. Then taking the quotient by $\langle (2,2) \rangle$ eliminates half of the copy $(k,k)$ (namely the even numbers). So what remains is all elements of $\mathbb Z \oplus \mathbb Z$ minus all elements of the form $(2n,2n)$?
Feb
24
comment Counting roots of polynomial inside $S^1$
@DanielFischer Yes, thanks. I was 99% sure it wasn't but... you (=I) never know. : ) I think I understand: It looks like if $a_n$ are the zeros of $p$ then the fraction ${p' \over p}$ expands as a sum $\sum_n {k_n \over z-a} $. Computing the integral of the sum yields $\sum_n k_n$ because $\oint {1 \over z-a}dz$ around $a$ is $1$. It's pretty neat.
Feb
24
comment Counting roots of polynomial inside $S^1$
@DanielFischer That $q$ in your first comment is not the same $q$ as in the question, right?
Feb
22
comment Concatenation with continuous function is entire
Thank you for your comment. I'm still a bit confused. I thought $\varphi$ was the linear operator $A_x$ here and $A_x$ is the derivative of $f$ at $x$. But then we'd have $\varphi = A_x = f'$ at $x$.Which of the equalities here I thought were true does not hold?
Feb
22
comment Concatenation with continuous function is entire
Did you mean $f'(a) =\varphi$ for all $a \in A$? (in the last sentence before the third quotation) Or perhaps $f'(a) =\varphi(a)$?
Feb
21
comment Continuity of the derivative
@Etienne Thank you for your comment. Indeed a seemingly not very thought out comment of mine, I'm sorry about that. Especially if one thinks about real functions where the linear map that is the derivative becomes the tangent!
Feb
21
comment Continuity of the derivative
@Etienne I am not sure if I understand your question correctly. If one considers $f: \mathbb C \to \mathbb C$ as a function from $f: \mathbb R^2 \to \mathbb R^2$ then if $f$ is (complex) differentiable at a point $x$ its derivative at $x$ is given by the Jacobian of $f$. But in finite dimension every linear map is continuous which would yield what I understand you are asking. But I think I misunderstand your question or I am probably making a fundamental mistake in what I wrote.
Feb
19
comment A closed ideal in a commutative Banach algebra $C(X)$
@DanielFischer I went ahead and made your comments into an answer. Hope it's ok.