35,170 reputation
12443
bio website jungenschaft-hohenstaufen.de
location Berlin, Germany
age 33
visits member for 3 years, 3 months
seen Oct 3 at 5:27

Aug
8
comment Riemann integrability of continuous function defined on closed interval
It's understandable and correct.
Aug
8
comment Find an equivalent to $(p\lor q) \to (p \lor r)$
Any equivalent? How about $(q \lor p) \to (r \lor p)$? Please explain in more detail what you want and what you have done.
Aug
8
comment Separability of the space of bounded uniformly continuous functions
Passing through the Stone-Cech compactification does not help in general, as the latter is not metrizable in general, and hence $C(\beta X)$ fails to be seperable (note, that for $K$ compact Hausdorff $C(K)$ is separable iff $K$ is metrizable).
Aug
7
comment Limit Question: $\lim_{x \rightarrow a} \frac{3}{x}$. What does the “a” represent?
This not an answer to the question asked in my humple opinion. The OP does not asked for the value of the limit, but for the "role" of $a$ ...
Aug
7
comment How to find rank of linear operator T on inner product space
You can $Tv = (v,\beta)\gamma$. When is this a multiple of $v$? (Note that it is always a multiple of $\gamma$!
Aug
7
comment Does “uniformly isolated” imply closed?
@goblin Better? Rephrased it a little.
Aug
7
comment Closed form for recurrence relation
The characteristic roots are (according to maple) \begin{align*} x_1 &= \frac{\bigl(108 + 12\sqrt{69}\bigr)^{2/3} + 12}{6\cdot \bigl(108 + 12\sqrt{69}\bigr)^{1/3}} \\ x_2 &= \frac{\bigl(108 + 12\sqrt{69}\bigr)^{2/3}(-1 + \sqrt{3}i) - 12(1 + \sqrt3 i)}{12 \cdot \bigl(108 + 12\sqrt{69}\bigr)^{1/3} }\\ x_3 = \overline{x_2} &= \frac{\bigl(108 + 12\sqrt{69}\bigr)^{2/3}(-1 - \sqrt{3}i) - 12(1 - \sqrt3 i)}{12 \cdot \bigl(108 + 12\sqrt{69}\bigr)^{1/3} } \end{align*}
Aug
7
comment How to find adjoint of linear operator T on inner product space V
The inner product is linear, $(\alpha,\beta)$ is just a scalar, let for a momoent $a := (\alpha, \beta) \in \mathbb C$. Then $$\bigl((\alpha, \beta)\gamma, \delta\bigr) = (a\gamma, \delta) = a(\gamma,\delta) = (\alpha,\beta)(\gamma, \delta) $$
Aug
7
comment convex weak* sequentially closed subset of a separable Banach space implies weak* closed
@niki Added something. Does this help?
Aug
7
comment Space of bounded functions vs. bounded space of functions.
@BenSouthworth If $\sup_{f \in B} \sup_{x\in \mathbb R} |f(x)|$ is bounded, all $f \in B$ are bounded functions. But on the other hand, all $f$ being bounded does not suffice! Let $B = \{f_n \mid n \in \mathbb N\}$, where $f_n \colon x \mapsto n$ is the constant-value-$n$-function. Then each $f_n$ is bounded, but $B$ is unbounded.
Aug
4
comment Statement regarding mean value theorem for harmonic functions
By your comment above, the inner integral equals $2\pi u(z_0)$
Aug
4
comment Statement regarding mean value theorem for harmonic functions
What is the form you know?
Aug
4
comment Efficiency in vector translation by matrix instead of vector
Exactly.${}{}{}$
Aug
1
comment Show that $\overline{A} \cup \overline{B} = \overline{A \cup B}$.
You are correct.
Aug
1
comment Singularities in Complex Analysis
Each singularity has it is own order. As for $\sin'(n\pi) \ne 0$ for each $n$, $\sin$ has first order zeros at $z = n\pi$, so $z/\sin z$ has first order poles at $z = n\pi$, $n \ne 0$, and a removable singularity at $z = 0$ (here the numerator is zero too and $z/\sin z \to 1$, $z \to 0$).
Aug
1
comment Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A \times B$ is closed in $X \times Y$.
up to some typos ... you want $y \not\in B$ not $x\not\in B$ (third line and sixth line)... fine
Aug
1
comment Sigma-additive measure on algebra of sets
What exactly do you want? Restrict your favourite measure to your favourite sub-algebra of its domain ...
Aug
1
comment Normal subgroup of a group
$K$ is normal, so for $h \in H$, $Kh = hK$. Does this help?
Aug
1
comment functional analysis operators
$U$ is not a map to $\mathbb R$? Do you want the codomain be $L^2([0,1])$?
Aug
1
comment Spaces where all singletons are closed
As I wrote, in general, more than "finite" cannot be achieved.