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Feb
7
revised If a group has no subgroups other than the identity and itself, then it is finite and is of prime order
deleted 151 characters in body
Feb
7
answered If a group has no subgroups other than the identity and itself, then it is finite and is of prime order
Feb
7
answered Proving function is infinitely differentiable
Feb
7
comment Proving function is infinitely differentiable
You should be able to prove this with the induction: the $n$th derivative of $h$ exists and is given by a composition of a smooth function and $h$.
Feb
4
comment Example of Lipschitz functions
Try any smooth function whose derivative is $> \varepsilon$ for some $\varepsilon > 0$.
Feb
3
comment Subspace of a weakly sequentially complete is weakly sequentially complete
To transport a quick summary: subspaces are strongly closed and convex, thus weakly closed. It is a general fact that a sequentially closed subset of a sequentially complete space is sequentially complete, in whatever topology.
Feb
3
comment Further generalising Holder's inequality
@math The inequality proven works for measurable functions.
Jan
31
comment Is the negation of a non-theorem a theorem?
I see. Thank you for clarifying.
Jan
31
comment Is the negation of a non-theorem a theorem?
You need a $\forall \text{ groups } G$ in front of your first formula, and so one of the two is true. In any case, if you specify which sets the $x$ and $y$ are being drawn from, you see that one of the two is true.
Jan
31
answered Determing the number of possible March Madness brackets
Jan
31
comment Exists a uniformly convex norm on Banach space satisfying certain condition?
This is Brezis FA problem 3.30 it appears for those wondering. Indeed your progress works. Of interest is the following inequality: $$| \frac{x+y}{2} |^2 \leq \frac{1}{2} |{x}|^2 + \frac{1}{2} |{y}|^2 - \delta$$. For each $M>0$, there exists a $\delta$ such this holds for all $x,y$ with norm less than $M$. I may post a complete proof here in the coming days.
Jan
28
comment For a convex function, why does lower semicontinuity imply weak lower semicontinuity?
For a general Banach space, weak sequential lower semicontinuity is not equivalent to weak lower semicontinuity. If you further assume $[f \leq \lambda]$ is bounded for all $\lambda$ and the dual is separable (or something resembling this), this should be enough.
Jan
28
comment For a convex function, why does lower semicontinuity imply weak lower semicontinuity?
That convex and strongly closed sets are weakly closed is a consequence of a geometric form of the Hahn-Banach theorem. That $[f \leq \lambda]$ is convex is a consequence of convexity of $f$. That $[f \leq \lambda]$ is closed is strong l.s.c. of $f$.
Jan
28
comment For a convex function, why does lower semicontinuity imply weak lower semicontinuity?
The set $[f \leq \lambda] $ is convex and strongly closed. Then it is weakly closed.
Jan
21
comment Question surrounding Exercise 3.12 of Brezis, function is convex and l.s.c. for the weak* topology.
The space is not assumed locally metrizable, so proving weak-$\ast$ sequential lower semi-continuity is insufficient.
Jan
20
comment How to determine if the sequence $\sum \limits_{n=1}^{\infty} \frac{\cos^2n}{n(n+1)} $ is absolutely convergent?
Hint: $\lvert \cos^2 n \rvert \leq 1$.
Jan
19
comment Confusion about $\text{Aut}(\mathbb{Z}/3) \simeq \mathbb{Z}/2$
The binary operation of $\text{aut}(G)$ is function composition by convention.
Jan
19
comment How does the pigeonhole principle intuitively suggest incorrect computations of probability?
You have a 100% chance of having the same result twice, but in half of these cases you throw out the first toss in the process of finding these two. Thus there is an extra factor of $1/2$ you are missing.
Jan
18
comment Which functions are $C^k$, but not $C^{k+1}$ on $ \mathbb{R} $
Perhaps an obstacle for a satisfying answer: "most" functions that are $C^k$ are not $C^{k+1}$. We can give certain properties of functions in this set, but our classifications must be limited in specificity to statements that apply to most $C^k$ functions.
Jan
18
comment Which functions are $C^k$, but not $C^{k+1}$ on $ \mathbb{R} $
If a function's $k$th derivative is identically $0$, then the function will be a polynomial, recovered by integration. I think the question as initially asked is compelling.