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Feb
6
accepted Continuous map from $L^r(\Omega)$ to $L^s(\Omega)$.
Feb
6
comment Continuous map from $L^r(\Omega)$ to $L^s(\Omega)$.
Ah, I was misreading the fact that the map is actually $x \mapsto g(x,\phi(x))$! Thank you.
Feb
6
asked Continuous map from $L^r(\Omega)$ to $L^s(\Omega)$.
Jan
29
awarded  Socratic
Jan
28
accepted Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
Jan
28
comment Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
Also lovely. Same question, though, any issues with what I've written?
Jan
28
comment Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
Will be happy to accept your answer after comment
Jan
28
comment Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
Ah, lovely. Any issues with what I've written?
Jan
28
revised Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
added 38 characters in body
Jan
28
comment Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
@Daga it's not "=" but if $x^2<1$, $x<1$...fixed
Jan
28
asked Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$
Jan
26
comment Convergence of Series of a.e. finite measurable functions
Much Appreciated
Jan
26
comment Convergence of Series of a.e. finite measurable functions
I'd be happy to give a +1 if you could elaborate on how these constants give convergence, and why they exist
Jan
26
comment Convergence of Series of a.e. finite measurable functions
@Giovanni I was looking at this problem too. Could you give a small hint as to how to proceed using BC?
Jan
25
accepted $\int_{\Omega} |f_n-f||f_n| \, d \mu \to 0$ if $f_n \in L^1(\Omega)$, $f_n \to f$.
Jan
25
comment Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.
@ForgotALot No it's definitely wrong, i was confusing the statement of the claim necessary for it to be true. See kobe's comments. Thanks for your help.
Jan
25
comment Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.
yeah, i made a counterexample a few moments ago. How silly of me. Though, I can't quite see how to go about this otherwise. I tried using FL, but I didn't know the above theorem. Thanks for your time.
Jan
25
accepted Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.
Jan
25
comment Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.
@ForgotALot I agree that there is an issue.
Jan
25
comment Given that $f_n \to f$ in $L^1(\Omega)$, $\mu(\Omega )=1$ and $ \|f_n\|_2^2 \leq M$, show $ \|f\|_2^2 \leq M$.
@ForgotALot Or how your example refutes the claim that if $$\int_{\Omega} f \, d\mu \leq \int_{\Omega} g \, d \mu $$ then $f \leq g$ a.e.