# Tag Info

## Hot answers tagged functional-analysis

6

Mesurable for the Lebesgue measure? Simple : they are continuous, and you can show that every continuous function is mesurable

2

You should check the Lawrence Evans' book about PDEs. More precisely, you can check its fifth chapter. There, you will see that, under some really general assumptions on the open set $\Omega \subset \mathbb{R}^n$, you have that $C^{\infty}$ is dense on $W^{k,p}$. About the other question, you generally have that $C^{\infty}_c$ is not dense on the ...

2

HINT: The Cauchy-Schwartz Inequality reveals that \begin{align} \left|Af(x)\right|^2 &= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\ &\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy \end{align} And thus, the square of the operator norm is \begin{align} ||A||_2^2&=\sup_{f\in \mathscr{L}^2} ... 1 By the inequality |Af| \leq A|f|, we may assume that f is non-negative. Then from the Tonelli's theorem (a.k.a. Fubini's theorem for non-negative functions),\| Af \|_2^2 = \int_0^1 \int_0^1 f(y)f(z) \left( \int_0^1 \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}} \right) \, dydz $$and we may try to estimate the following function$$ k(y, z) = \int_0^1 ...

1

There are only two such equivalence relations. In one, all three elements are equivalent. In the other, 1~2 (as required) but 3 is not equivalent to any element (other than itself).

1

Here are some references that might help: Wikipedia page: https://en.wikipedia.org/wiki/Nash%E2%80%93Moser_theorem "An Inverse Function Theorem in Frechet Spaces" by Ivar Ekeland: https://www.ceremade.dauphine.fr/~ekeland/Articles/InverseFunctionTheorem.pdf "On the Nash-Moser Implicit Function Theorem" by Lars Hormander: ...

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