Question

Product of an $L^\infty$ function and an $L^1$ function is integrable

score 1.3k · Answer 1 · 2026-04-14 02:18:49

1.3k

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Product of an $L^\infty$ function and an $L^1$ function is integrable

Published on 14 Apr 2026 - 2:18

score 124 · Answer 2 · 2026-04-12 19:20:55

124

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Factorization of $L^{p}$ spaces

Published on 12 Apr 2026 - 19:20

score 1.1k · Answer 3 · 2026-04-14 16:58:34

1.1k

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Why is $f(x) = \sin(x)$ an element of $L^2(-\pi, \pi)$ not $L^2(a,b)$

Published on 14 Apr 2026 - 16:58

score 1.1k · Answer 4 · 2026-04-11 01:02:27

1.1k

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What is the cardinality of $L^p(\mathbb R)$, $1 \le p < \infty$?

Published on 11 Apr 2026 - 1:02

score 974 · Answer 5 · 2026-04-11 23:23:26

974

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Weak convergence and trace operator

Published on 11 Apr 2026 - 23:23

score 625 · Answer 6 · 2026-04-10 21:43:27

625

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Is the convolution pointwise bounded?

Published on 10 Apr 2026 - 21:43

score 510 · Answer 7 · 2026-04-13 17:47:41

510

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Why is $\mathfrak{L}^1$ not a vector space of functions that take infinity as a value

Published on 13 Apr 2026 - 17:47

score 55 · Answer 8 · 2015-08-10 03:21:11

55

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"$f\ge 0$, $g>0$, $fg\in L^1(\mathbb{R})$ and $g\notin L^1(\mathbb{R})$" implies "$f$ is integrable over $[-r,r]$"?

Published on 10 Aug 2015 - 3:21

score 1.9k · Answer 9 · 2026-04-12 21:43:48

1.9k

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schauder basis for $\ell_\infty$

Published on 12 Apr 2026 - 21:43

score 399 · Answer 10 · 2026-03-30 03:56:22

399

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If $X_n\to X$ in $L^p$, then $E(X_n)^p \to E(X)^p$

Published on 30 Mar 2026 - 3:56

score 1.2k · Answer 11 · 2015-08-11 04:07:29

1.2k

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Assume that $f_n\to f$ in measure and $\sup_n\|f_n\|_{L^p(E)}<\infty$ for some $p>1$. Prove that $f_n$ converges to $f$ in $L^1$ norm.

Published on 11 Aug 2015 - 4:07

score 224 · Answer 12 · 2026-04-13 11:33:16

224

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Doob's inequalities for not necessarily right-continuous martingales

Published on 13 Apr 2026 - 11:33

score 99 · Answer 13 · 2015-08-12 12:00:28

99

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Functions by which one can multiply elements of $L^1_{\text{loc}}$

Published on 12 Aug 2015 - 12:00

score 442 · Answer 14 · 2015-08-12 13:01:47

442

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Taking the limit $\lim_{p\rightarrow \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p$

Published on 12 Aug 2015 - 13:01

score 35 · Answer 15 · 2026-04-13 02:10:26

35

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The convolution of compact $L^1$ weighted function is still $L^1$

Published on 13 Apr 2026 - 2:10

List Question

Product of an $L^\infty$ function and an $L^1$ function is integrable

Factorization of $L^{p}$ spaces

Why is $f(x) = \sin(x)$ an element of $L^2(-\pi, \pi)$ not $L^2(a,b)$

What is the cardinality of $L^p(\mathbb R)$, $1 \le p < \infty$?

Weak convergence and trace operator

Is the convolution pointwise bounded?

Why is $\mathfrak{L}^1$ not a vector space of functions that take infinity as a value

"$f\ge 0$, $g>0$, $fg\in L^1(\mathbb{R})$ and $g\notin L^1(\mathbb{R})$" implies "$f$ is integrable over $[-r,r]$"?

schauder basis for $\ell_\infty$

If $X_n\to X$ in $L^p$, then $E(X_n)^p \to E(X)^p$

Assume that $f_n\to f$ in measure and $\sup_n\|f_n\|_{L^p(E)}<\infty$ for some $p>1$. Prove that $f_n$ converges to $f$ in $L^1$ norm.

Doob's inequalities for not necessarily right-continuous martingales

Functions by which one can multiply elements of $L^1_{\text{loc}}$

Taking the limit $\lim_{p\rightarrow \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p$

The convolution of compact $L^1$ weighted function is still $L^1$

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