498 reputation
211
bio website
location
age
visits member for 2 years, 2 months
seen 3 hours ago

Jul
2
awarded  Curious
Jun
3
awarded  Yearling
May
26
comment Why is G a cycle? Graph G with n vertices, m edges. “(∀v ∈ V(G) : δ(v) = 2 ⇒ G is Hamiltonian) is true because G is a cycle”.
Yes. See this question: math.stackexchange.com/questions/390030/…
May
26
comment Why is G a cycle? Graph G with n vertices, m edges. “(∀v ∈ V(G) : δ(v) = 2 ⇒ G is Hamiltonian) is true because G is a cycle”.
Your example fails because neither $a$ nor $c$ have degree 2.
May
23
accepted Haar measure of point sets
May
23
asked Haar measure of point sets
May
15
comment Let $f$ is a continuously differentiable function then what can we say about the function $\frac{1}{f}$ ?
The function $f(x) = x$ defined on the reals is continuously differentiable (smooth, even), but $1/f$ is not even continuous everywhere, let alone continuously differentiable.
Apr
30
comment Riemann Hypothesis Proof - legit?
The references are priceless too. There are four in total: one is a link to the Russian Wikipedia page on "Numbers", another one is a link to a page with the first 50 million prime numbers, and the last two are papers written by the author.
Mar
21
comment Norm of operator between two $L^p$ spaces?
Thanks for the reply; I made an edit to clarify what I was asking about the relation between the $|| \cdot ||_{p \to q}$ norm and the $ || \cdot ||_p$ and $ || \cdot ||_q$ norms.
Mar
21
revised Norm of operator between two $L^p$ spaces?
deleted 7 characters in body
Mar
21
asked Norm of operator between two $L^p$ spaces?
Mar
8
revised Block Diagonalisation of 4x4 Matrix
Removed representation theory tag
Mar
8
suggested suggested edit on Block Diagonalisation of 4x4 Matrix
Feb
22
answered Mathematical Induction on Matrix Sequence
Nov
28
awarded  Popular Question
Oct
7
revised Failure of Doob-Dynkin lemma in general measurable spaces
added 16 characters in body
Oct
2
comment If $f(x)$ and $g(x)$ are Riemann integrable and $f(x)≤h(x)≤g(x)$, must $h(x)$ be Riemann integrable? (Repost)
Without the assumption, certainly not: on your domain of definition, let $f$ be the constant function equal to $0$, $g$ the constant function equal to $1$, $h$ the indicator function on the rationals.
Sep
28
revised Failure of Doob-Dynkin lemma in general measurable spaces
edited tags
Sep
28
asked Failure of Doob-Dynkin lemma in general measurable spaces
Aug
23
comment Deciding whether this map is well defined.
$f(1 + 20\mathbb{Z}) = 1 + 15\mathbb{Z}$. But $1 + 20\mathbb{Z} = 21 + 20\mathbb{Z}$ and $f(21 + 20\mathbb{Z}) = 21 + 15\mathbb{Z} = 6 + 15\mathbb{Z} \neq 1 + 15\mathbb{Z} $.