# Guillermo

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 Feb16 comment For what values of $p>0, \quad \int^{1}_{0} \frac{x}{\sin{(x^{p})}} \operatorname d\!x$ converges? There isn't any oscillation near $0$ to prevent the integral to blowing up as you make $t$ apporach $0$. So my guess is that this exists whenever $0< p <2$ and diverges for all $p \geq 2$. That this blows up for $p\geq 2$ can easily be proved using the estimate $\sin(x) \leq x$. That it converges for $00$. Can you show your work? Jan29 comment Problem in walter rudin RAC So you want to prove that $f$ is not in $L^1$, or is it that $f'$ is not in $L^1$? Jan29 answered Help with inequality estimate, in $H^1$, Jan22 answered Multiplicative operator from L1 to L1 is given by an L_inf function Jan21 comment Integral convergence and weak convergence @JohnJack oooh sorry, I misread. I don't know why it would converge to that, there must be some property of $a$ I am missing. Jan20 comment Integral convergence and weak convergence @JohnJack It's the same argument: $\nabla z \in L^p(\Omega)$, so the first term converges by the weak convergence hypothesis. Jan18 comment Show $(h_1\mu_1)\otimes\ldots\otimes (h_n\mu_n)=h(\mu_1\otimes\ldots\otimes\mu_n)$ (and that this is a probability measure) it looks correct to me Jan18 comment Integral convergence and weak convergence @JohnJack I have updated my answer with a proof of this fact. But this is probably in Brezis' book: amazon.com/…. Jan18 revised Integral convergence and weak convergence added further explanations Jan18 comment This function is continuous and has the following estimation? For the usual topology of $C(\mathbb{R})$, if $f$ is continuous in $[-n,n]$ for all $n$, then $f$ must be continuous since the topology is just that of pointwise continuity. Also, under your hypotheses, $\phi(f) > \sum 2^n = \infty$