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55

It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's Linear Algebra, 4th Edition, which has an extensive 8 part "Hint." Here's an edited sequence of hints, following theirs: First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle ...


48

I sent Professor Purdy an email. I asked him what he recalled about the incident. With his permission I've copied his correspondence below. Dear Jacob, Yes, I was there, and I'm the one who told the story to Paul Hoffman, who then included it in his book "The man who loved only numbers." The 30 page proof was written by Jack Bryan just ...


45

Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit ...


43

Let me ask you a dual question: I am a mathematics student in set theory, why don't category theory students do set theory? I find it strange that most books on category theory have only a naive handling of set theory. Now let me answer your question. Category theory is an impressive tool for abstraction, but analysis is not always in need for abstraction - ...


41

[...] Grothendieck, with his background in functional analysis, must have been familiar with Stone's work in that field. cited by Qiaochu reveals a gap in the knowledge of Grothendieck's work (which is of course nothing to be blamed for, given its vastness and ramifications in so many fields) but it also misses a link to one of Grothendieck's beautiful ...


35

Robert's and joriki's examples are of course nice and explicit, but you can get examples on any subset of $\mathbb{R}^n$ with infinite measure. Here's how: Take a function $f$ that is in $L^p$ but not in $L^q$ for $q \gt p$ (on the unit ball $B$ around zero, say). Now take a sequence $x = (x_n)$ that is in $\ell^p$ but not in $\ell^q$ for $q \lt p$ (there ...


34

Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of "nice" functions on real values - e.g. bounded and having continuous derivatives. The action of operators $L^h$ on this space has a semigroup structure: $$ L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = ...


33

The essential idea of many transforms is to change the basis in the space of functions with the hope that in the new basis the problem will simplify. Let me give a finite-dimensional example. Suppose we have a $2\times2$ matrix $A$ and we want to compute $A^{1000}$. Direct approach would not be very wise. However, if we first diagonalize $A$ as $PA_dP^{-1}$ ...


32

Suppose $T: L^\infty \to \ell^\infty$ is your isomorphism. Looking at this coordinatewise, this corresponds to a bounded sequence $\phi_n$ of bounded linear functionals on $L^\infty$ such that 1) there is $\epsilon > 0$ such that $\max_n |\phi_n(f)| \ge \epsilon \|f\|_\infty$ for all $f \in L^\infty$ 2) For every bounded sequence $t_n$ of reals there ...


31

It seems to me that you've got the things quite right. I restrict attention to real or complex vector spaces. (Edit: But see the update.) Given a scalar product $\langle \cdot, \cdot\rangle$ we get a norm by setting $\|x\| = \sqrt{\langle x,x \rangle}$ by an application of the Cauchy-Schwarz inequality. Conversely, a normed vector space structure comes ...


31

Yes, you are right. It was Jack Bryant, not Jack Brian. He might have retired by now. By the way, even before Erdos came to town, it was generally agreed that there must be a proof that was shorter than 30 pages, but not a two liner! Professor Don Allen talked to Erdos the next day to see if he could help with the research problem that had generated this ...


29

There are three main ways of interpreting the Fourier transform. Decomposition relative to eigenfunctions of the Laplacian On $\mathbb{R}^n$, the plane waves $E_\xi(x) = \exp( i \xi\cdot x)$ can be interpreted as generalized eigenfunctions of the Laplacian. That is, let $$ \triangle = \sum_{i=1}^n \left(\frac{\partial}{\partial x_i}\right)^2 $$ then ...


28

Yes, if $(\Omega, \Sigma, \mu)$ is a (complete) $\sigma$-finite measure space then $(L^{\infty}(\Omega,\Sigma,\mu))^{\ast}$ is the space $\operatorname{ba}(\Omega, \Sigma,\mu)$ of all finitely additive finite signed measures defined on $\Sigma$, which are absolutely continuous with respect to $\mu$, equipped with the total variation norm. The proof is ...


27

There is a Theorem of Wielandt which asserts that if $A$ is any normed algebra, complete or not, we can't express $I = 1_{A}$ in the form $xy - yx$. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that $xy -yx = I$. I claim that $xy^{n} - y^{n}x = ny^{n-1}$ for all $n \in \mathbb{N}$. We have the case $n = 1.$ Suppose ...


26

Here's a sketch of a proof. Let $\sigma(x)$ denote the spectrum of $x$. Then $\sigma(xy)\cup\{0\} = \sigma(yx)\cup\{0\}$. On the other hand, $\sigma(1+yx)=1+\sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $\sigma(xy)$ is unbounded. But every ...


26

Since $1/x$ is the border case in both directions, the most promising candidate would be a modified version of $1/x$ that just converges but won't converge if you nudge it ever so slightly. We have $$\int_2^\infty \frac1{x\log^2x}\mathrm dx=\left[-\frac1{\log x}\right]_2^\infty=\frac1{\log2}\;,$$ whereas $$\int_2^\infty ...


26

The closed unit ball of $c_0$ has no extreme points. The closed unit ball of $c$ has many extreme points, such as $(1,1,\ldots)$. Since the property of being an extreme point is preserved by isometries, $c$ and $c_0$ are not isometrically isomorphic.


25

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism. But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.


25

You are talking about what is often called a measure-linear mean or limite médiale in the sense of Mokobodzki. $\newcommand{mypart}[1]{\unicode{x2014;}\text{ #1 }\unicode{x2014;}}$ Edit: Given my recent revisiting the literature on this topic, it seems like medial limit (when thought of as an ultrafilter on $\mathbb{N}$) or medial mean (when thought ...


25

Considering the fact that you have only had one undergraduate course in analysis and will be taking an actual functional analysis class, I don't think you actually want to self-study functional analysis. It would be much more useful for you to bulk up on your linear algebra review your real analysis Functional analysis is, for a large part, linear ...


24

Let $V$ be a vector space over the field $\mathbb{F}$. A norm $$\| \cdot \|: V \longrightarrow \mathbb{F}$$ on $V$ satisfies the homogeneity condition $$\|ax\| = |a| \cdot \|x\|$$ for all $a \in \mathbb{F}$ and $x \in V$. So the metric $$d: V \times V \longrightarrow \mathbb{F},$$ $$d(x,y) = \|x - y\|$$ defined by the norm is such that $$d(ax,ay) = \|ax - ...


24

Section 1 Let us begin with the following theorem. Theorem 1 Let $ G $ be a topological group. If $ G $ admits a Lie group structure, then this structure is unique up to diffeomorphism. Proof: Suppose that $ \mathcal{A}_{1} $ and $ \mathcal{A}_{2} $ are smooth structures (maximal smooth atlases) on $ G $ that make it a Lie group. Observe that the ...


24

The uniform boundedness principle of Functional Analysis is a very important application of the Baire Category Theorem. Added: (t.b.) See also Sokal's A really simple elementary proof of the uniform boundedness theorem for a proof without Baire.


23

Here's $\DeclareMathOperator{\ev}{ev}$ the answer if $(C[0,1], \|\cdot\|)$ is assumed to be complete. Consider the family $\mathcal{F} = \{\ev_{x}\}_{x \in [0,1]}$ of continuous linear functionals on $(C[0,1],\|\cdot\|)$. For each $f \in C[0,1]$ we have $\sup_{x \in [0,1]} |\ev_{x}(f)| \leq \|f\|_{\infty}$, so the family $\mathcal{F}$ is pointwise bounded. ...


23

No. First note that if $X$ is any set, the space of all functions $f: X \to \mathbb{R}$ with the topology of pointwise convergence is simply the space $\prod_{X} \mathbb{R}$ with the product topology. It is well-known that this space is metrizable if and only if $X$ is at most countable, see e.g. J. Dugundji, Topology, IX.7.1, p.189. However, if $X$ is ...


23

What is the spectral theorem for compact operators good for? Here are some examples. (I am ignoring the self-adjoint aspects, since they don't really play a role in the theorem. And it is valid for more general spaces than Hilbert spaces too, so I will also ignore that part, in the sense that I won't pay too much attention to whether my examples deal with ...


22

Fix $\delta>0$ and let $S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$ for $\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since $\mu(S_\delta)$ is finite and positive. This gives ...


22

Try $$f(x) = \frac{1}{x^{1/p} (\ln(x)^2+1)} \qquad \text{on} \qquad (0, \infty)$$


21

Clearly, vector spaces are the most general notion among these, because they can be defined over any field and all, Banach, Hilbert and Sobolev spaces are vector spaces. Banach spaces make sense over any normed field (e.g. $\mathbb{R}$ or $\mathbb{C}$). I've only encountered Hilbert spaces over the reals or the complex numbers so far. Every Hilbert space is ...


21

There are two questions here, in reality, I think. First, in brief, I am told by many people that I "do functional analysis in the theory of automorphic forms", and I certainly do find a categorical viewpoint very useful. Second, in brief, it is my impression that the personality-types of many people who'd style themselves "(functional) analysts" might be ...



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