List questions from lp-spaces

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Brezis Functional Analysis Exercise 8.10

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Weak convergence in $L^2$ of a sequence bounded in $H^1$ implies weak convergence in $H^1$

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If $f_n(X) \subset \{1,2, \ldots, 101\}$ for all $n$, then $f(x) \in \{1,2,\ldots,101\}$ for almost every $x \in X$.

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Prove that $L_p$-norm is non-convex, $p < 1$, $p \not = 0$.

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for infinite interval, $L^{\infty}$ convergence implies $L^{2}$?

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Sequence of functions that converges in L∞ but not L2?

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Hausdorff-Young inequality on T space

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How $\langle f,\phi\rangle_{L^2} = 0 \ (\le0)$ implies $f = 0 \ (\le0)$

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Show density in $\ell^2$

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Norms for $L^2$?

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Is the functional $I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $ weakly lower semicontinuous?

Published on 23 Feb 2026 - 10:04

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How is the dominated convergence theorem applied here?

Published on 26 Apr 2020 - 8:32

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$\int_0^1 e^{2x}dx - |\int_0^1 e^x \overline{\phi(x)}|^2= \int_0^1|e^x - \int_0^1 e^t \overline{\phi(t)}dt\phi(x)|^2dx$ in $L_2[0,1]$

Published on 26 Apr 2020 - 10:33

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Small in uniform measure cannot be too big in another measure.

Published on 26 Apr 2020 - 18:00

List Question

prove that a measurable function is in $L^{\infty}([0,1])$

Brezis Functional Analysis Exercise 8.10

Weak convergence in $L^2$ of a sequence bounded in $H^1$ implies weak convergence in $H^1$

If $f_n(X) \subset \{1,2, \ldots, 101\}$ for all $n$, then $f(x) \in \{1,2,\ldots,101\}$ for almost every $x \in X$.

Prove that $L_p$-norm is non-convex, $p < 1$, $p \not = 0$.

for infinite interval, $L^{\infty}$ convergence implies $L^{2}$?

Sequence of functions that converges in L∞ but not L2?

Hausdorff-Young inequality on T space

How $\langle f,\phi\rangle_{L^2} = 0 \ (\le0)$ implies $f = 0 \ (\le0)$

Show density in $\ell^2$

Norms for $L^2$?

Is the functional $I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $ weakly lower semicontinuous?

How is the dominated convergence theorem applied here?

$\int_0^1 e^{2x}dx - |\int_0^1 e^x \overline{\phi(x)}|^2= \int_0^1|e^x - \int_0^1 e^t \overline{\phi(t)}dt\phi(x)|^2dx$ in $L_2[0,1]$

Small in uniform measure cannot be too big in another measure.

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