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13 votes
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Functions whose product with every $L^1$ function is $L^1$

Yes, the answer to your question is positive. For any function $f\in L_1$, such that $f \not\in L_\infty$, we will explicitly find $g\in L_1$, such that $fg \not\in L_1$. Let us take arbitrary non-...
Jarosław Błasiok's user avatar
6 votes
Accepted

Is there a term for a countour integral that disregards direction?

That is the line integral of a scalar field. Consider a parameterization $\gamma(t)$, $a<t<b$ of the contour $C$, then $$\int_C f(z) \, dz=\int_a^b f(\gamma(t))\gamma'(t) \, dt,$$ while $$\...
Carlo Beenakker's user avatar
5 votes

Functions whose product with every $L^1$ function is $L^1$

I think this also a direct consequence of the uniform boundedness principle (see https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Corollaries): $T_n \colon g \mapsto \int \, f g \mathbb{1}_{...
unwissen's user avatar
  • 533
2 votes
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Integrability in the product space can follow from a property of the Nemytskii operator?

Nemytskii operators on Lebesgue spaces are funny objects with a lot of implicit structure. In fact, if $\mathcal N_f$ maps $L^p(\Omega)$ into $L^q(\Omega)$, then $f$ must necessarily satisfy the ...
Hannes's user avatar
  • 2,300
2 votes

Is there a term for a countour integral that disregards direction?

In Calculus textbooks $$\int_\gamma f(z)|dz|$$ is called the line integral of the first kind, In arbitrary dimension it is frequently written as $$\int_\gamma f(x)ds,$$ where $ds$ is the length ...
Alexandre Eremenko's user avatar
1 vote

For what sets does the Lebesgue Differentiation Theorem hold in one dimension?

In chapter 2 of Federer's book, Geometric Measure Theory, it is developed the theory of Vitali's coverings. That theory is exactly the answer to your question
Luigi De Pascale's user avatar

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