TITLE: Adjoint of multiplication by $z$ in a Hilbert Space (Bergman space)
QUESTION [6 upvotes]: I am learning Hilbert space theory from Halmos' "Introduction to Hilbert space and the theory of spectral multiplicity".
While talking about understanding adjoints (p. 39), he calls special attention to this example, remarking that "its adjoint is not what at first it might appear to be":
Let $\mathfrak H$ be the set of analytic functions defined in the interior of the unit disk ($D$), square integrable with respect to planar Lebesgue measure. Then $\mathfrak H$ –called a Bergman space– is a Hilbert space with the inner product $$\langle f,g \rangle= \int_D f\bar g d\lambda = \int_D f(x+iy)\bar g (x+iy) dxdy .$$
In $\mathfrak H$, consider the multiplication by $z$ opertator $A$, i.e. $(Af)(z)=zf(z).$
In the typical $L^2$ I think $A^*$ would simply be multiplication by $\bar z$, but that ruins the differentiability of $f$, so in this case it must be some other thing.
I have thought that this multiplication operator works like the shift operator in sequence spaces if one identifies the function $f$ with its power series terms $(a_0, a_1, ..)$, mapping this sequence to $(0, a_0, a_1,..)$. I know that the usual right shift defined in $l^2$ has as adjoint the left shift when one considers the inner product $\langle \{a_n\},\{b_n\}\rangle = \sum a_n\bar b_n$, but I don't know how the inner product of $\mathfrak H$ would look like translated to the language of its corresponding sequence space, so this approach hasn't helped me much either.
How can I construct this adjoint operator?

REPLY [3 votes]: The idea with the sequence space is a good one. A very nice feature of $\mathfrak{H}$ is that its Hilbert space structure fits nicely with Taylor expansion about $0$: The functions $p_n \colon z\mapsto z^n$ are mutually orthogonal.
$$\begin{align}
\langle p_k, p_n\rangle &= \int_D z^k\cdot \overline{z^n}\,d\lambda\\
&= \int_0^1 \int_0^{2\pi} r^k e^{ik\varphi}\cdot r^n e^{-in\varphi} \,d\varphi\; r\,dr\\
&= \int_0^1 r^{k+n+1} \left(\int_0^{2\pi} e^{i(k-n)\varphi}\,d\varphi\right)\,dr\\
&= \begin{cases} \dfrac{\pi}{n+1} &, k = n\\\quad\vphantom{\dfrac12} 0 &, k \neq n. \end{cases}
\end{align}$$
Since they span a dense subspace, the scaled functions form a Hilbert basis. Let us denote
$$b_n(z) = \sqrt{\frac{n+1}{\pi}}\cdot z^n.$$
Then $(b_n)_{n\in\mathbb{N}}$ is an orthonormal basis of $\mathfrak{H}$, and the multiplication with $z$ in that basis becomes
$$\begin{align}
z\cdot f(z) &= z\cdot \sum_{n=0}^\infty c_n\cdot b_n(z)\\
&= z\cdot \sum_{n=0}^\infty c_n\sqrt{\frac{n+1}{\pi}}\cdot z^n\\
&= \sum_{n=0}^\infty c_n\sqrt{\frac{n+1}{\pi}}\cdot z^{n+1}\\
&= \sum_{n=1}^\infty c_{n-1} \sqrt{\frac{n}{\pi}}\cdot z^n\\
&= \sum_{n=1}^\infty c_{n-1}\sqrt{\frac{n}{n+1}}\cdot b_n(z),
\end{align}$$
so it's not just a shift of the corresponding sequence, but a shift together with a multiplication of the terms with the sequence $\sqrt{\frac{n}{n+1}}$.
That makes it easy to determine the adjoint in terms of the sequence of coefficients with respect to $(b_n)$. Translating that into the representation by the Taylor series and then to the function-level gives a basis-free characterisation of the adjoint (involving a definite integral).


TITLE: Regarding playing an infinite number of games that could last infinitely long amounts of time
QUESTION [2 upvotes]: So after watching the last Stanley cup game, a problem popped up in my head for which I have no solution. Say we have a game, like a hockey game, that has the possibility of going on forever. Of course, it does not usually go on forever, nor have we ever seen it go on forever. But we'd like to!
We also would like to see every single possible outcome that could happen in the game. Naturally, there are an infinite number of possible outcomes, as this is a mathematically pure game uninhibited by the inelegant rules of our physical realm. However, this does not stop us! We decide to create an algorithm to enumerate all possible outcomes. It's very simple, it proceeds as thus:
While True:
    Play a new game

Now, as we have already stated, one of the possible outcomes of playing a new game is having a game that runs on forever. According to the infinite monkey theorem, this is bound to happen at some point. But when it does, we will be unable to continue this algorithm, as it is impossible to say whether or not the game goes on forever before it, well, erm, goes on forever (halting problem).
So here's where I'm stuck. Does this mean it's impossible to enumerate all possible outcomes of playing this game? Is finding all possible outcomes undecidable (even though there are an infinite number of outcomes so finding all solutions is inherently impossible)?
Unfortunately, I do not have the education (yet!) to deduce the solution with proper vocabulary. I'm sure this problem has existed for a long while in some other form (I'm thinking halting problem, or somehow closely related), so this shouldn't be very mysterious. Thanks!

REPLY [0 votes]: The way to do it would be to interleave plays in the game. This is akin to how we can simulate all computations, even the ones that don't halt: run step one of the first computation, then step one of the second, then step two of the first, then step two of the second, then step one of the third...


TITLE: Cardinal arithmetic questions
QUESTION [4 upvotes]: I have problem to solve:
Let $a,b$ and $c$ be cardinal numbers. Prove that $a+b=b$, $b \le c$ implies $a+c=c$.
And trying to prove this I got couple questions:

For infinite cardinal $c$, is $c+c=c$ always? And how to prove it? (Is $c \cdot c=c$?)
From $a+b=b$ can I conclude that $a \le b$ since $a \ge b$ and 1. will imply $a+b=a$? I know that statement that every two cardinals are comparable leads to AC (we can use it when solving problems)

(I apologize if my questions are too simple, I tried to google it, but Google is not math-friendly tonight for me =D)
P.S. Is there solution without using AC?

REPLY [3 votes]: You asked how to prove (1). Assuming the axiom of choice, every cardinal is an aleph.
Let $\kappa=\aleph_\gamma$ for some $\gamma$. To prove that $\kappa+\kappa=\kappa$, note that every ordinal $\alpha$ can be written uniquely in the form $\alpha=\eta+n$, where $\eta$ is $0$ or a limit ordinal, and $n\in\omega$. Say that $\alpha$ is even if $n$ is even and odd if $n$ is odd. Let $K_0=\{\alpha\in\kappa:\alpha\text{ is even}\}$ and $K_1=\{\alpha\in\kappa:\alpha\text{ is odd}\}$. Then
$$\varphi_0:\kappa\to K_0:\alpha=\eta+n\mapsto\eta+2n$$
and
$$\varphi_1:\kappa\to K_0:\alpha=\eta+n\mapsto\eta+2n+1$$
are bijections, and $\kappa=K_0\sqcup K_1$, so
$$\varphi:2\times\kappa\to\kappa:\langle k,\alpha\rangle\mapsto\varphi_k(\alpha)$$
is a bijection from $2\times\kappa$ onto $\kappa$.


TITLE: Find marginal probability density function without the joint density function or the other marginal pdf
QUESTION [0 upvotes]: This is a question from exam review sheet. Please give me some guidance here. I do not know how can I find fY(y) without having information on f(x,y) or at least fX(x)?
Consider a random variable Y generated as follows. First select a value of X = x at random (uniform) over the interval (0,1). Then select a value of Y = y at random (uniform) over the interval (0,x). Find the probability density function fY(y).
Thank you.

REPLY [2 votes]: You know $X\sim U(0,1)$. Further, you know conditional on $X=x$, that $Y\sim U(0,x)$.
That is the conditional density of $Y$ given $X=x$ is $f(y|x)=\frac{1}{x} \mathbb{1}_{0<y<x}$.
The unconditional density can then by found by the law of total probability, i.e.
$$f_Y(y)=\int_0^1 f(y| x)f_X(x)dx,$$
Can you finish up from this?


TITLE: What is the base measure in measure theory?
QUESTION [4 upvotes]: I see the term "base measure" used frequently about measures. I do not completely get what that exactly means:
Some examples are:

Let $\cal F$ be the space of all probability density functions with
  respect to a base measure $\nu$

What is the base measure?
Sometimes when a probabilistic function is integrated,

the dx is called a base measure.
   $$\int_{\cal X}  ....  dx$$

Can someone explain in simple words or refer me to a simple reference to read about "base measures".

REPLY [3 votes]: When I see the word "base measure", the first thing that comes to me is the Radon-Nikodym theorem (and the Radon-Nikodym derivative).
The Radon-Nikodym theorem states the following.

Consider a measurable space $(X,\Sigma )$ on which two $\sigma$-finite measures $\mu$ and $\nu$ are defined. If $\nu$ is absolutely continuous with respect to $\mu$ (i.e., $\mu (A) = 0$ implies $\nu (A) = 0$ for all measurable $A \subseteq X$), then there exists a $\Sigma$-measurable function $f:X \rightarrow [0,\infty )$, such that for any measurable set $ A \subseteq X$, $$ \nu (A)=\int _{A}f\,d\mu.  $$

In addition, if two functions $f_1$ and $f_2$ satisfies the same property, then $$\mu ( \cup_{A \subseteq X, A\text{ measurable} } \{ A : f_1 (A) \neq f_2 (A) \} ) = 0.$$
In words, we say such $f$ is $\mu$-almost everywhere unique. Because of this uniqueness one often write $ f = \frac{ d \nu }{ d \mu } $ (provided that $\nu$ is absolutely continuous with respect to $\mu$). In such cases, people often call $\mu$ the base measure.
For me (who mostly works in Euclidean spaces and rarely does research in measure theory), I think of the Lebesgue measure and assume that a density function can be properly written out when I hear the word base measure.