Suppose that for each $f \in L^p (\mathbb{R})$, the improper integral $$\int_{\mathbb{R}} f(x) \varphi(x) dx$$ converges. Show that $\varphi \in L^q (\mathbb{R}).$
My idea is that by the Holder inequality, $$\left| \int_{\mathbb{R}} f(x) \varphi(x) dx \right| \le \bigg\{ \int_{\mathbb{R}} \left| f(x) \right| ^p dx \bigg\}^{\frac{1}{p}} \bigg\{ \int_{\mathbb{R}} \left| \varphi(x) \right| ^q dx \bigg\}^{\frac{1}{q}}$$
The left side converges, thus both factors on the right also converge. Therefore $$\int_{\mathbb{R}} \left| \varphi(x) \right| ^q dx < \infty$$ and consequently, $\varphi \in L^q (\mathbb{R})$.
However the flaw in the proof is that it doesn't guarantee that the left-hand side is less than $\infty$
Hmm, this is somewhat harder than showing that $(L^p)^* = L^q$, since we don't know yet that the map $f\mapsto \int_{\mathbb{R}} f\phi$ is bounded on $L^p$, do we?
This inspiration isn't the best but closed graph theorem is kind of one of theorems that lets you infer boundedness of an operator based on knowing a sort of "soft" fact
@0celo7 according to André a bunch of the results are just plain wrong cause they'd resolve some of the major problems in MCF but i havent seen why they fail yet
For $p=2$, that operator is symmetric. A symmetric operator is closable. It has to equal its closure since it's defined everywhere. But a closed operator defined everywhere is continuous by CGT.
Maybe something like that works in the other cases.
you might ask which operator I'm talking about, and I don't have a good answer lol
We have the subset $A=\{(x,y)\in \mathbb{R}^2 : y=\sin\frac{1}{x}, x\in (0,\frac{1}{\pi})\}$. I want to determine the boundary of $A$.
Do we maybe check for that where a convergent sequence of the set converges?
It holds that $(a_n, b_n)\rightarrow (a,b)\iff a_n\rightarrow a$ and $b_n\rightarrow b$.
We have two cases: if $a\neq 0$ and if $a=0$.
If $a\neq 0$: Let $(a_k,b_k)\in M$. We have that $a_k\neq 0, \forall k\geq k_0$ since $a_k\rightarrow a\neq 0$. We also have that $b_k=\sin\frac{1}{a_k}\rightarrow b\in [-1,1]$.
Since it is a graph the interior of A is empty, because if we consider a ball around a point of the graph, we will see that there is a point on the ball that doesn't belong to the graph, right?
hello, im having trouble with using mathjax in this chat. I added the links on this page (math.ucla.edu/~robjohn/math/mathjax.html) to my bookmarks but it doesnt work. the tex on the installation page is displayed but not here in the chat. im not very familiar with html and so on
No, I mean, it's the graph of a continuous function @MaryStar
$x\mapsto \sin {1\over x}$ is a continuous function over $\Bbb R^*_+$
Which means the restriction of the graph to any compact is closed
ie $\forall \epsilon \gt 0$, $A_\epsilon = \{(x,\sin{1\over x}), x\in [\epsilon, {1\over\pi}]\}$ is closed and of empty interior, and is its own boundary
Now something different happens at $0$, because it cannot be continued as a continuous function
As a matter of fact, due to the periodicity of $\sin$, you can prove the the boundary is $A \cup \{({1\over \pi}, 0)\} \cup \left([0;1]\times\{0\}\right)$
And more generally, if $f$ is a periodic continuous function over $\Bbb R$, defining $A_f = \{(x, f({1\over x})), x\in \Bbb R_+^*\}$, you have $\overline {A_f } = A_f \cup ( \{0\}\times f(\Bbb R))$
^^^ $\{0\}\times [0;1]$, not $[0;1]\times \{0\}$
Meaning if $f$ isn't constant, $x\mapsto f(1\over x)$ cannot be continued by a continuous function at $0$
In the (undercrowded) Quantum Computing SE beta site someone asked about Explicit Lieb-Robinson Velocity Bounds. I found the question worthy and offered my first bounty on it... which is still unclaimed. Maybe some people from the math community are up for that? There are no answer yets, so any helpful contribution has a very good chance to claim the bounty...
If V^* is a vector space dual to V, are the basis elements e^i = \delta_{ij} linear? For example, can I do e^i(e_i + e_j) = e^i(e_i) + e^i(e_j)? I assume not because the LHS e^i(e_i + e_j) = 0, but RHS is 1?
Why does this show that those three functions are linearly independent? I mean why is it justified to take different values of x and then derive a=b=c?
@Silent They're linearly independent if there exist constants a, b, and c such that $ax+be^x+c\sin(x)=0$. Since a, b, and c are constants, they don't depend on $x$, so we can plug in different values of $x$.
@Silent You could try $x=-1$, $0$, and $1$ and get a system of three linear equations in $a$, $b$, and $c$, and try to show (using determinants or elimination) that it doesn't have any nontrivial solutions.
Some day I will read again why the wronskian plus suitable intersection of intervals can control the behaviour of uncountably many points of the functions
It will be interesting to see what nontrivial set of functions f,g,h are linearly dependent over the whole real line (that will require countably many intersections)
@AkivaWeinberger, Does there exist unique matrices $P,L,D,U$ for any invertible matrix $A$, such that $PA=LDU$, where $P$ is permutation matrix, $L$ is lower triangular matrix with all diagonal entries $1$, $D$ is a diagonal matrix and $U$ upper triangular matrix with all diagonal entries $1$. ? (I know existence.)
Hi, I would like to know: What kind of graph that can be split into small number of trees. For example: planar graph could be split into 2 or 3 using k-tree. Is there any other special kind of graph.
Hi, I'm having some troubles understanding how the dimension of a projective variety $X \in \mathbb{P}^n$ is defined.
I know that if $Y$ is an affine variety, then the dimension of $Y$ is equal to the dimension of its underlying topological space. Does this definition also work per a projective variety?
@konoa If $X \subset \Bbb P^n$ is your projective variety, it has a corresponding homogeneous ideal $\mathcal{I} \subset k[X_0, \cdots, X_n]$ that defines it. Algebric dimension of $X$ is one less than the Krull dimension of $k[X_0, \cdots, X_n]/\mathcal{I}$
The reason we take one less is because the common zero set of $\mathcal{I}$ defines the cone over $X$ in $\Bbb A^{n+1}$, which is one dimension more than that of $X$.
(That's one of the many ways to define dimension, I'm sure)
You can think of it as the dimension of $X$ as a complex manifold (if its smooth), so for example Riemann surfaces are 1-dimensional complex manifolds (but of course they are surfaces when viewed as a real manifold)
-- In general in algebraic geometry you define the dimension of a variety /scheme as the dimension of its underlying topological space (equipped with the zariski topology) - where by dimension you mean the supremum of the length chain of irreducible closed subsets. Note that this is a stupid definition if your space is Hausdorff - you can check that with this definition with \R…
I have two measure spaces $(A,\Sigma_1,\mu)$ e $(B,\Sigma_2,\nu)$ and I consider one of their products $(A\times B,\mu\times\nu,\Sigma_1\otimes\Sigma_2)$ where $\Sigma_1\otimes\Sigma_2$ is the $\sigma$-algebra generated by sets of the form $F\times G$ for $F\in\Sigma_1$ and $G\in\Sigma_2$ (we can take $\mu$ and $\nu$ to be $\sigma$-finite so that the product measure is unique if that makes things easier)
In general it may be possible to extend $\mu\times\nu$ to a $\sigma$-algebra bigger than $\Sigma_1\otimes\Sigma_2$, for example the product measure might not be complete even if the two factors are
Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction?
Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{...
@0celo7, i really can't understand this: Suppose $U$ is a subspace of a vector space $V$, then there exists another vector space $W$ such that basis of $U$ together with basis of $W$ give basis of $V$. Then why for $v\in V$, there is unique $u\in U$ and $w\in W$ such that $u+w=v$?
@DarkVampiricAbstractArtist, ok, sorry which topic is that from? Like I understand how int v is v^2 (altho shouldn't it be v^2/2?) but how do I get the 25?
@DarkVampiricAbstractArtist, no, I get that, I'm just missing the rule that says to add lower limit to the integrand when lower limit is a constant and upper limit is a variable - or is the formulation different?
In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if
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@Abcd, no its fine now. I mean, I still don't know the formal rule that says the whole of LHS is over 2, not just v^2, and I'd like to. But I see what procedure to follow.
@user17629 I suspect that there is a way of obtaining something like a Laplacian on a space with uniform structure (which is weaker than metrizable), though I have no idea how it might work. I doubt that you could weaken the hypotheses beyond that. Why?
@0celo7 I think it's more interesting to work with the completion then. In this case I can't say that a measurable set every section measurable so it is meaningful to ask the question above
Hi all; I'm stuck on a basic problem: If Ms. Simon starts her drive at 6:30 a.m., she candrive at her average driving speed with no trafficdelay for each segment of the drive. If she starts herdrive at 7:00 a.m., the travel time from the freewayentrance to the freeway exit increases by 33% due toslower traffic, but the travel time for each of the othertwo segments of her drive does not change.
Based on the table, how many more minutes does Ms. Simon take to arrive at her workplace if she starts her drive at 7 AM than if she starts her drive at 6:30 AM?
@NicholasRoberts I think this is true - say m \ne n. Since Q(root(n),root(m)) contains Q(root(m))+Q(root(n)), it suffices to show that both are degree 4 extensions of Q. I think you can check by hand that the min poly of root(m) + root(n) is not of deg 2, and of course it satisfies a deg 4 poly. On the other hand Q(root(n),root(m)) is a degree 2 extension of Q(root(m)).
the point is the so one of sqrt(n) + sqrt(m)'s conjugates is sqrt(n) - sqrt(m) (you can check: the minimal polynomial of this dude is (x^2 - n)^2 - m), so both sqrt(n) and sqrt(m) is contained in Q(sqrt(n), sqrt(m))
Okay, let $A$ be a self-adjoint algebra of complex continuous functions on a compact set $K$. Suppose $A$ vanishes nowhere and separates points on $K$.
Let $R$ denote the uniform closure of $A$. Why is it that if $f= u+iv$ and $u \in R$ and $v\in R$, then $f \in R$?
it does come down to the question of whether $P=LQU$ with $L$ lower-triangular, $P$ and $Q$ permutation matrices, and $U$ upper-triangular can occur with $P\neq Q$
@0celo7 I like how it gets straight to the interesting stuff without wasting time, the author managed to fit an incredible variety of topics in the same book. Interestingly the measure they use as a counterexample is even finite so I don't think there's any hope of getting a "measurable slices implies measurable" kind of result under nice assumptions
Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction?
Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{...
