Question

If $f^p\in L^1([0,1])$ it's bounded a.e.

score 98 · Answer 1 · 2026-04-19 20:16:24

98

Views

If $f^p\in L^1([0,1])$ it's bounded a.e.

Published on 19 Apr 2026 - 20:16

score 774 · Answer 2 · 2026-04-14 04:03:41

774

Views

Characterization of weak convergence in Lp

Published on 14 Apr 2026 - 4:03

score 38 · Answer 3 · 2026-04-17 21:27:35

38

Views

If $f \in L^{\infty}(\Bbb R^d) \cap C_u(\Omega)$, then $\rho_{\epsilon} * f \to f$ uniformly over $\Omega$

Published on 17 Apr 2026 - 21:27

score 23 · Answer 4 · 2026-04-16 10:36:56

23

Views

$f(t)=t^{2r-2p-1} (e^{2t}-e^{-2t})^{-1/2}\in L^{1}(0, \infty)$?

Published on 16 Apr 2026 - 10:36

score 109 · Answer 5 · 2026-04-19 04:09:07

109

Views

$X$ be Banach , $T:X \to \mathcal l ^{\infty}$ be linear , $(Tx)_n$ the $n$-th term of $T(x)$;$f_n(x)=(Tx)_n$ ; if each $f_n$ is bdd then so is $T$?

Published on 19 Apr 2026 - 4:09

score 1k · Answer 6 · 2026-04-12 19:53:08

1k

Views

Showing that a subspace of $L^p$ is closed

Published on 12 Apr 2026 - 19:53

score 109 · Answer 7 · 2026-04-17 10:03:30

109

Views

$\forall g\in L^2(\Omega)$ exists $g_n\in H_0^1(\Omega)$ and $\epsilon>0$ s.t $g_n(x)\to g(x),\,a.e$ and $|g_n(x)|\leq |g(x)|+\epsilon$

Published on 17 Apr 2026 - 10:03

score 413 · Answer 8 · 2026-04-15 19:39:56

413

Views

If an $H^1$ function vanishes on a set of positive measure, its $L^2$ norm is controlled by the gradient

Published on 15 Apr 2026 - 19:39

score 894 · Answer 9 · 2026-04-12 10:54:07

894

Views

$L^{\infty}$ convergence for random variable

Published on 12 Apr 2026 - 10:54

score 142 · Answer 10 · 2026-04-15 13:08:54

142

Views

Compactness Sobolev embedding for even functions on $\mathbb{R}$.

Published on 15 Apr 2026 - 13:08

score 99 · Answer 11 · 2026-04-07 00:25:14

99

Views

If $f_j\to f$ in $L^1(\Bbb R^n)$ then $Tf_j\to Tf$ in $L^{1,\infty}(\Bbb R^n)$

Published on 07 Apr 2026 - 0:25

score 86 · Answer 12 · 2026-04-07 00:25:14

86

Views

If $||g_j-g||_{1,\infty}\to0$, then $||g_j||_{1,\infty}\to||g||_{1,\infty}$

Published on 07 Apr 2026 - 0:25

score 1.6k · Answer 13 · 2026-04-19 12:48:54

1.6k

Views

Show there is a continuous isomorphism $l_{\infty}\rightarrow \left(l_1\right)^* $

Published on 19 Apr 2026 - 12:48

score 609 · Answer 14 · 2026-04-09 15:31:53

609

Views

The Lebesgue Fundamental Theorem of Calculus

Published on 09 Apr 2026 - 15:31

score 226 · Answer 15 · 2026-04-16 18:24:51

226

Views

If $f \in L^2[-\pi, \pi]$, then $f \in L^1[-\pi, \pi]$ and$\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$

Published on 16 Apr 2026 - 18:24

List Question

If $f^p\in L^1([0,1])$ it's bounded a.e.

Characterization of weak convergence in Lp

If $f \in L^{\infty}(\Bbb R^d) \cap C_u(\Omega)$, then $\rho_{\epsilon} * f \to f$ uniformly over $\Omega$

$f(t)=t^{2r-2p-1} (e^{2t}-e^{-2t})^{-1/2}\in L^{1}(0, \infty)$?

$X$ be Banach , $T:X \to \mathcal l ^{\infty}$ be linear , $(Tx)_n$ the $n$-th term of $T(x)$;$f_n(x)=(Tx)_n$ ; if each $f_n$ is bdd then so is $T$?

Showing that a subspace of $L^p$ is closed

$\forall g\in L^2(\Omega)$ exists $g_n\in H_0^1(\Omega)$ and $\epsilon>0$ s.t $g_n(x)\to g(x),\,a.e$ and $|g_n(x)|\leq |g(x)|+\epsilon$

If an $H^1$ function vanishes on a set of positive measure, its $L^2$ norm is controlled by the gradient

$L^{\infty}$ convergence for random variable

Compactness Sobolev embedding for even functions on $\mathbb{R}$.

If $f_j\to f$ in $L^1(\Bbb R^n)$ then $Tf_j\to Tf$ in $L^{1,\infty}(\Bbb R^n)$

If $||g_j-g||_{1,\infty}\to0$, then $||g_j||_{1,\infty}\to||g||_{1,\infty}$

Show there is a continuous isomorphism $l_{\infty}\rightarrow \left(l_1\right)^* $

The Lebesgue Fundamental Theorem of Calculus

If $f \in L^2[-\pi, \pi]$, then $f \in L^1[-\pi, \pi]$ and$\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$

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