tchao
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No, this is not true. Take the series $\nu_n=\frac{1}{n}$.

I think you already wrote down everything you need to solve your problem. About measurability. I assume that we are speaking about the filtration $\mathcal F_t$ that is generated by the given ...

Yes, it is true. Since the series $\sum_{n=1}^\infty |a_n|^2$ converges to zero, the sequence $S_N:=\sum_{n=1}^N |a_n|^2$ must converge to zero. But we have: $S_N\geqslant S_{N-1}\geqslant 0$, ...

The statement is true if your function is continuous. Any continuous real-valued function on a compact set takes its infinum (and supremum) at some point in this set. If $g(z)$ is continuous, $|g(z)|$ ...

This sentence is the negation of the statement that $a_n\to -\infty$. If $a_n\to -\infty$, then for every $\beta\in\mathbb R$, there exists $n_0\in\mathbb N$ such that for all $n>n_0$ we have $a_n&... View answer Accepted answer 1 votes 1) Yes. 2) Note that on the right-hand side of your estimate you have$\lVert \nabla u \rVert_{L^2}^2\leq \lVert u \rVert_{H_0^1}^2$and$\lVert u \rVert_{L^2}^2\leq \lVert u \rVert_{H_0^1}^2$, see ... View answer Accepted answer 1 votes The formula from your comment is correct. Observe that for sets$C, D, E, F$we have $$C-(D\cup E)=(C-D)\cap (C-E),$$ $$(C-D)\times E=C\times E - D\times E,$$ and $$(C\times D)\cap (E\times F)=(C\... View answer 1 votes Because$$\frac{(t-1)(t^{n-1}+...+1)}{(t-1)(t^{m-1}+...+1)}=\frac{t^{n-1}+...+1}{t^{m-1}+...+1},$$and all powers of t converge to 1 as t goes to 1. There are n summands in numerator (the n-... View answer 1 votes The probability \mathbb P(A\cap B) is minimal (actually, zero) if A and B are complementary events, like A having heads and B having tails when throwing a coin once. Note that these events ... View answer Accepted answer 1 votes It is$$E[X|Y]=\sum_{y \in \mathcal Y}\Big[\sum_{x \in \mathcal X}x\cdot P(X=x|Y=y)\Big]\cdot \chi_y(Y),$$where \chi_y(Y)=1 if Y takes the value y and zero otherwise. In that way, you indeed ... View answer 1 votes We can derive separate formulas for (x_{2k}) and (x_{2k+1}). For even indices we have: x_2=0 and x_{2k+2}=\frac{1}{2}+\frac{x_{2k-2}}{2} for k>1. One can easily show by induction that ... View answer Accepted answer 1 votes I would rather say that if A is diagonalizable, it is also "jordanizable", and its Jordan form J is exactly D, the corresponding diagonal matrix. Think of every diagonal element of D as of a ... View answer Accepted answer 0 votes Recall that B_t is normally distributed with mean 0 and variance t. Then by the "law of the unconscious statistician",$$\mathbb E(\sin(5B_t))=\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}\sin(... View answer Accepted answer 0 votes With the hints of Sergei Golovan: Consider conditional expectation$ E(|X+Y|^p|\ X)$and apply Jensen's inequality to it: $$E(|X+Y|^p|\ X )\geq |E(X+Y|\ X)|^p.$$ Moreover, since$X$and$Y$are ... View answer 0 votes (a) Your value of$a$seems to be incorrect, but you use correct relations. (b) The distribution function is an increasing function such that$\lim_{t\to -\infty }F(t)=0$,$\lim_{t\to +\infty }F(t)=...
You should introduce two more functions $v$ and $w$, so that your system of three equations does indeed contain three unknown functions. Then you specify that $v$ and $w$ are, respectively, the first ...