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Does $L^p$ convergence preserve the $L^1 $ norm of the sequence

If $\mu(U)$ is finite - yes, just by the fact $\|f\|_1 = \int f \cdot \operatorname{sign}(f)\, d\mu = \|f\|_p \cdot \mu(U)^\frac{p}{p - 1}$, so convergence in $L_p$ implies convergence in $L_1$. ...
mihaild's user avatar
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3 votes
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Missing argument in the proof of the Levy-Khintchine representation .

Hopefully there is a more concise answer. I will prove much more than required, namely that $f_n(x) dx$ converges in distribution towards $\delta_0(dx) + f(x) dx$ with $f$ decreasing, and that $f_n \...
Thomas Lehéricy's user avatar
2 votes
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weak convergence and pointwise implies $L_p$ convergence

In general, the statement in the title of the OP is false. In $((0,\infty),\mathscr{B}(\mathbb{R}),m)$, $m$ is Lebesgue measure, define $f_n(x)=\frac1{x-n}\mathbb{1}_{(1,\infty)}(x-n)$. $f_n\...
Mittens's user avatar
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Galerkin method for Poisson's equation

Due to Theorem 1 in 6.5.1, there is an orthonormal basis of $L^2$ consisting of smooth functions $w_k\in H_0^1$ which satisfy $-\nabla^2 w=\lambda_k w_k$ where $\lambda_k\rightarrow \infty$ and $\...
Kadmos's user avatar
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2 votes
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Weak convergence of degenerated random variable with finite support. Does it work?

For $x<0$, $F(x)=F_n(x)=0$. For $x>0$ and $n$ large enough, $1/n<x$ so $F(x)=F_n(x)=1$. So the convergence of cdfs holds on all continuity points of $F$.
Will's user avatar
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1 vote

Exercise 2.8 from Billingsley

Let $d$ denote the metric on $S$. Also, let $K = \overline{\{ x_n : x \in \mathbb{N} \}}$ denote the closure of the image of $(x_n)$. Then the lemma below will ensure that the sequence satisfying the ...
Sangchul Lee's user avatar
2 votes
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Exercise 2.8 from Billingsley

When $S$ is separable, perhaps more direct proof can be given as follows. Fix $x\in\operatorname{supp}(P)$. For any open neighborhood $V$ of $x$ let $f\in C_b(S)$ such that $0\leq f\leq 1$, $f(x)=1$ ...
Mittens's user avatar
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2 votes
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Moment existence under weak $L^1$-convergence

No. Pick any $f$ that is continuous, in $L^1(\mathbb{R}^d)$, and that does not admit $q$-th moment. (Say, $f: \mathbb{R} \to \mathbb{R}$, $f(x) = \frac{1}{x^2}$ when $|x| \geq 1$ and $f(x) = 1$ ...
David Gao's user avatar
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