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asked
Is there $r>0$ such that the norm $[f]_r:=\sup_{n \ge 1} \|1_{B(y_n, r)} f\|_{L^p}$ is equivalent to $\|f \|:=\sup_{y \in Y}\|1_{B(y,1)}f\|_{L^p}$?
asked
Is the embedding $i: (L^p_\text{loc} (Y), \| \cdot \|_{L^p_\text{loc}}) \to (L^0(Y), \hat \rho)$ continuous or Borel measurable?
comment
The set of measurable functions together with convergence in measure is a completely metrizable abelian topological group
@NateEldredge You are right! It should be $\hat{\rho}(1/n, 0) = 1/n$. However, it seems I have found a new counter-example. Let $f_n (x) =f(x) =x \in \mathbb R$ and $\lambda_n = \frac{1}{n}$ and $\lambda =0$. Then for all $\delta>0$, $$ \mu (|\lambda_n f_n - \lambda f| > \delta) = \mu ( \{x \in \mathbb R : |x| > n \delta\}) = +\infty. $$
revised
The set of measurable functions together with convergence in measure is a completely metrizable abelian topological group
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asked
The set of measurable functions together with convergence in measure is a completely metrizable abelian topological group
asked
Can we modify this extended pseudometric such that its convergence is equivalent to that in measure?
comment
Is this metric on the space of $\mu$-measurable functions complete?
@IosifPinelis Thank you so much for your indication! I'm happy that the answer is affirmative.
comment
Is this metric on the space of $\mu$-measurable functions complete?
@GeraldEdgar Yess, I use Bochner measurability. Have you sent me the wrong link? It's "How to ask a good question".
asked
Is this metric on the space of $\mu$-measurable functions complete?
accepted
Is there a real/functional analytic proof of Cramér–Lévy theorem?
revised
Is there a real/functional analytic proof of Cramér–Lévy theorem?
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asked
Is there a real/functional analytic proof of Cramér–Lévy theorem?
accepted
Are there $f,h$ such that $h$ is Lipschitz, $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(h(t), x)$?
asked
Are there $f,h$ such that $h$ is Lipschitz, $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(h(t), x)$?
comment
Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$?
Thank you so much for your counter-example!
accepted
Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$?
revised
Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$?
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asked
Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$?
accepted
Let $g$ be the heat kernel. Are there constants $C_1, C_2>0$ such that $\frac{g(t_1, \cdot)}{t_1} \le C_1 \frac{g(C_2 t_2, \cdot)}{\sqrt{t_2}}$?
revised
Let $g$ be the heat kernel. Are there constants $C_1, C_2>0$ such that $\frac{g(t_1, \cdot)}{t_1} \le C_1 \frac{g(C_2 t_2, \cdot)}{\sqrt{t_2}}$?
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