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## Hot answers tagged lp-spaces

4

You can repeat the usual proof of completeness of $L^p$. Namely, let $\{f_n\}$ be Cauchy; by choosing a subsequence we may assume that $\|f_{n+1}-f_n\|_{L^{p,r}}<2^{-n}$. Define $$g=|f_1|+\sum_j|f_{j+1}-f_j|,$$ this is a measurable function because it is a sum/limit of positive measurable functions. Now \begin{align} \left\||f_1|+\sum_{j=1}^n|f_{j+1}-...

3

Yes, in fact you can do it explicitly. Simply find $g$ with $\| g \|_{L^q} = 1$ such that $fg = C|f|^p$, hence $\int f g = C \| f \|_{L^p}^p$. Then check that the requirement $\| g \|_{L^q}=1$ sets $C=\frac{1}{\| f \|_{L^p}^{p-1}}$. In the special case $p=2$, you have the intuitive result $g=\frac{f}{\| f \|_{L^2}}$.

3

You can't do that, because $\ell^1$ is not closed in $\ell^\infty$. The sequence $x=\left(\frac1n\right)_{n\in\mathbb N}$ belongs to $\ell^\infty\setminus\ell^1$. But, if, for each $n\in\mathbb N$, $x_n$ is the sequence$$1,\frac12,\frac13,\ldots,\frac1n,0,0,\ldots,$$then $(\forall n\in\mathbb N):x_n\in\ell^1$ and $\lim_{n\to\infty}x_n=x$ in $\ell^\infty$.

2

For non-negative real random variables $Y$ we have $EY=\int_0^\infty P(Y>t)dt$, which you could apply to $Y=|f|^p$. So in your first example, where $\mu(f>n)=O(n^{-a})$ you can check if $\sum_n n^{-a/p}<\infty$, and so on.

2

Case 1: $q<\infty$. From $$|f(x)|^r \leq |f(x)|^q, \qquad x \in \{|f|>1\}$$ and $$|f(x)|^r \leq |f(x)|^p, \qquad x \in \{|f| \leq 1\},$$ it follows that \begin{align*}\int |f(x)|^r \, dx &= \int_{\{|f| \leq 1\}} |f(x)|^r \, dx + \int_{\{|f| > 1\}} |f(x)|^r \, dx \\ &\leq \int |f(x)|^q \,dx + \int |f(x)|^p \, dx < \infty.\end{align*} Case ...

1

For $p<2$ set $f = \frac{h}{\|h\|_2}$, where $h = g^{\frac p{2-p}}$. Then $\|Tf\|_p = \|g\|_{\frac{2p}{2-p}}$, so $\|T\|$ coincides with the latter value (and is even attained). For $p=2$ choose some $x_0\in [0,1]$ such that $|g(x_0)| = \|g\|_\infty$. Let $\epsilon > 0$. Then there exists $\delta > 0$ such that for $x$ in the $\delta$-neighborhood $... 1 Yes, this is usually known as Riesz representation theorem. We have that for any$\omega \in (L^p)^*$there is a unique$g\in L^q$such that$\|\omega\|=\|g\|_q$. That is$\|g\|_q=\sup_{\|f\|_p=1}|\omega(f)|$; but under the dual pairing$|\omega(f)|=|\langle f, \omega \rangle |= |\langle f, g \rangle |$. 1 Partial Answer $$(\rho \ast f)(x)=\int_{-1}^1 e^{-{1 \over {} 1-y^2}}(x-y)dy$$ $$=x\int_{-1}^1 e^{-{1 \over {} 1-y^2}}dy-\int_{-1}^1 ye^{-{1 \over {} 1-y^2}}dy$$ The second integral is zero because the integrand is an odd function. Also the first integral is a constant. I am not sure if you can calculate it explicitly. 1 That's almost the integral you want to look at, but the power is not quite right. If you transform into polar coordinates in$\mathbb{R}^n$, you'll get $$\int\limits_{B_1(0)}\frac{1}{|x|^q}\, dx=\int\limits_{S^{n-1}}\int\limits_0^1\frac{1}{r^q}r^{n-1}\, drd\sigma(\omega),$$ where$d\sigma$denotes the surface measure on the$S^{n-1}$and$\omega\in S^{n-1}$... 1 The space$H$is the closure with respect to the norm induced by$(\cdot,\cdot)_1$of the linear hull of$\epsilon$. Thus, for all elements in$H$, the induced norm is finite. This implies that$x\in L^2$and$x'\in L^2$for all$x\in H\$.

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