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Jun
29
comment Hausdorff-Young / Restriction Inequality
To make it even more explicit: take $f_i$ to be "frequency translations" of one fixed $f_0$ which has compact support in frequency space.
Jun
29
comment Hausdorff-Young / Restriction Inequality
Unfortunately, I have neither the time nor energy to hold your hand through every single step of the argument. (I really suggest you discuss this with your tutor or professor, since from your comments you are missing some very basic techniques in analysis.) So one last-ditch effort: to prove an inequality $\|Tf\|_{L^q} \leq C \|f\|_{L^p}$ is false, the usual way is to construct a sequence $f_i$ such that $\|f_i\|_{L^p} = 1$ and $\|Tf_i\|_{L^q} \nearrow \infty$. So see if you can construct such an $f_i$ using exactly the ingredients that I gave above.
Jun
26
comment Is :$\frac{\Bbb d}{\Bbb d x}$ a chaotic operator in infinite-dimensional Hilbert space?
You have given us an abstract Hilbert space, but you have not told us how $d/dx$ acts on elements of the Hilbert space. Are you thinking about $H = L^2(\mathbb{R})$ or something? In which case $d/dx \not\in B(H)$ so is trivially not a chaotic operator by your definition.
Jun
25
comment Conditions under which a function vanishing on the boundary belongs to $H_0^1$
$H^1_0$ is usually defined as the closure of $C^\infty_0$ (or equivalently $C^1_0$) in the $H^1$ norm. Functions in $C^\infty_0$ or $C^1_0$ have compact support.
Jun
25
comment Hausdorff-Young / Restriction Inequality
1. Let $K$ be any compact set, on the complement $\mathbb{R}^d \setminus K$, there are not mutual inclusions between $L^p$ and $L^q$ for $p\neq q$. 2. Let $h:\mathbb{R}^d \setminus K \to \mathbb{R}$ be a function with $\inf h > M > 0$. Then the mapping $f\mapsto hf$ for $f\in L^p$ is either unbounded, or has operator norm at least $M$. 3. Try to prove the statement for the case the density of $\mu$ is $(1 + |x|^2)^\sigma$ for $\sigma > 0$. The general case follows not too more difficultly.
Jun
25
awarded  Nice Answer
Jun
24
answered Two different trigonometric identities giving two different solutions
Jun
24
comment “Commutation” of parallel transport with covariant derivative and Riemann curvature tensor
Note that for the second question, there is no need for $X,Y,Z$ to actually be vector fields along $c$. Your question as posed only depends on the pointwise values of $X,Y,Z$, so we can consider the equivalent problem by extending those vectors to vector fields along $c$ by parallel transport.
Jun
24
comment “Commutation” of parallel transport with covariant derivative and Riemann curvature tensor
For 2, the answer is obviously no. You are equivalently asking the following question: if $X, Y, Z$ are vector fields parallel transported along $c$, is $R(X,Y)Z$ a vector field that is parallel transported along $c$? Taking the covariant derivative with respect to $c$ you see that this is only the case if $\nabla_{c'} R = 0$.
Jun
24
comment “Commutation” of parallel transport with covariant derivative and Riemann curvature tensor
The RHS of your first equation makes no sense. $P_{t_1,t_2}(Y(t_1))$ is a single vector in $T_{c(t_2)}M$, and not a vector field along $c$, so it is meaningless to take the $c'$ covariant derivative of it.
Jun
23
comment Conditions under which a function vanishing on the boundary belongs to $H_0^1$
$u_\delta$ is certainly not necessarily smooth. It equals $u$ when $|u(x)| > 2$ so if $u$ is not smooth there neither will $u_\delta$. The key thing about the hint is not that $u_\delta$ is smooth, but that $u_\delta$ has compact support in $\Omega$.
Jun
23
comment Conditions under which a function vanishing on the boundary belongs to $H_0^1$
Side comment: it is generally a good idea to ask two tangentially related questions as separate ones on this website, rather than as one question with two distinct parts.
Jun
23
comment Hausdorff-Young / Restriction Inequality
The sum comes from the example in the back of my mind (sorry for not being clear) when $\mu$ has a density that can be decomposed as the sum of a $L^r$ function and a $L^\infty$ function. (If the density is not bounded near $\infty$ then you expect to lose derivatives on the physical side and cannot generally expect the sort of restriction inequality you want to hold; so when I see your question I think of the case where the density is bounded outside a compact set and within the compact set is only $L^r$.)
Jun
22
comment Is $ (((\sqrt{2})^ \sqrt{2})^ \sqrt{2})^{\cdots} $ an irrational number?
Voting to re-open based on @rajb245's comment
Jun
22
comment Overview of nonlinear analysis, differential equations (ODE and PDE), dynamical systems, and mathematical physics, and their relationships
Can't help you with that. Any narrowing will be a different question from what you have asked, and I can't read your mind (sorry) well enough to guess where you would prefer the emphasis be placed.
Jun
22
revised Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
added 1765 characters in body
Jun
22
comment Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
Jun
20
awarded  Favorite Question
Jun
19
comment Hausdorff-Young / Restriction Inequality
The the density of $\mu$ is in some $L^r_{\text{loc}}$, then Holder's inequality is basically the best you can do there; the right hand side will be a sum of $L^p$ and $L^{s}$ where $s$ is determined by $r$ and Hausdorff Young. You can check that by a scaling argument.
Jun
19
answered $U$ bisected by all hyperplanes $\implies$ $U$ symmetric?