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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 7 months |
| seen | 1 hour ago | |
| stats | profile views | 942 |
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May 14 |
comment |
Dominated convergence application? Do you mean you don't know how to prove the hint or how to use the hint or both? |
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May 14 |
revised |
boundedness of an operator added 31 characters in body |
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May 14 |
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boundedness of an operator @ougao: I am terribly sorry for a serious mistake in the last version of my answer. I didn't realize the mistake until I was occasionally involved in this post. The current proof of boundedness of $\|T\|$ is essentially borrowed from here. I Hope I have fixed all the problems now. |
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May 14 |
revised |
boundedness of an operator added 400 characters in body |
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May 14 |
comment |
Proving that $ \int \left| f-g \right|~d\mu = 2\int_{A_0} (f-g)~d\mu$ You are welcome. |
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May 14 |
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Show that the linear operator $(Tf)(x)=\frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ satisfies $\|T\|\leq 1$. @robjohn: Well down! Maybe you should remind the OP of emphasizing $\|T\|=1$ in the description of question, to distinguish it from the linked question on $\|T\|\le 1$. |
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May 14 |
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Show that the linear operator $(Tf)(x)=\frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ satisfies $\|T\|\leq 1$. @robjohn: Sorry, it was my mistake. I am not sure if $\|T\|=1$ or not now. |
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May 13 |
revised |
Property of partial traces added 365 characters in body |
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May 13 |
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Prove that $c^3f(c)+c f(c)\ge1$ @Jimmy_Jp: Please don't be discouraged by those questions. Many questions will become easier to be handled once you get more experienced. Some other questions may be so tricky or artificial that it doesn't matter if we cannot solve them. |
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May 13 |
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Prove that $c^3f(c)+c f(c)\ge1$ @Jimmy_Jp: Thank you for your appreciation! |
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May 13 |
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Prove that $c^3f(c)+c f(c)\ge1$ @Jimmy_Jp: You are welcome. Just in case that you are new to this website, you may choose to accept an answer to your question if you wish. |
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May 13 |
answered | Prove that $c^3f(c)+c f(c)\ge1$ |
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May 13 |
comment |
Map Quadrant Conformally onto the Unit Disc and find $|g'(1+i)|$. @Tsotsi: Firstly, the map $z\mapsto z^2$ maps the first quadrant comformally to the upper half plane; secondly, a linear fractional transformation maps lines to circles/lines. |
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May 13 |
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How to prove that there exists a $z_0 \in U_{1} [0]$ such $ \prod_{k=1}^{n} |z_0 - a_k | \geq 1 $ for $a_1, \dots , a_n \in U_{1} [0] $? A nice more elementary argument than mine, +1! A comment: be careful about $f_0$, which has no lower bound. Nevertheless, this gap could be fixed by playing some trick with the definition of $f$. |
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May 13 |
revised |
slowness of growth of polynomials added 471 characters in body |
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May 13 |
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$2^n-3^m=1 , m,n \in \mathbb N =?$ @Arjang: Thank for your reply. I respect your decision. |
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May 13 |
answered | Property of partial traces |
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May 13 |
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Map Quadrant Conformally onto the Unit Disc and find $|g'(1+i)|$. You are welcome. Would you like to post an answer by yourself? |
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May 13 |
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Map Quadrant Conformally onto the Unit Disc and find $|g'(1+i)|$. Why do you set $g(z) = \frac{az + b}{cz + d}$? You may try $g(z) = \frac{az^2 + b}{cz^2 + d}$ instead. |
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May 13 |
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When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$? Yes, your example definitely works. It was my mistake to say $0$ in the support of $f_0$. Actually I want a litte more: $f_0^{(k)}(0)\ne 0$ for some $k\ge 2$ to exclude bump functions, and now I see achille hui has cleared my doubts. Thank you for your replies. |
