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Aug
16
asked Asymptotic for $\int_{\mathbb{R}}e^{ik(sin x - x)}f(x)\, dx$
Aug
16
accepted Where is $\int_{-\infty}^{\infty}\frac{f(y)}{\sqrt{(y - x_{1})^{2} + x_{2}^{2}}}\, dy$ continuous?
Aug
6
comment Where is $\int_{-\infty}^{\infty}\frac{f(y)}{\sqrt{(y - x_{1})^{2} + x_{2}^{2}}}\, dy$ continuous?
Can't I have a convergent integral even if the function diverges? For example, if I'm intergrating over a ball of radius R in $\mathbb{R}^{2}$, $\int_{|x| \leq R}\frac{1}{|x|}\, dx = \int_{0}^{2\pi}\int_{0}^{R}\frac{1}{r}r\, dr\, d\theta = 2\pi R$?
Aug
6
comment Where is $\int_{-\infty}^{\infty}\frac{f(y)}{\sqrt{(y - x_{1})^{2} + x_{2}^{2}}}\, dy$ continuous?
Thanks for the comment! After you're initial comment, I had an idea, I was wondering if it works? We have that the integral is equal to $g(x) = \int_{\mathbb{R}^2}\frac{f(y)1_{y_{2} = 0}}{|x - y|}\, dy$. Then to prove continuity of this, it is enough to show that $\int_{\mathbb{R}^2}|f(y)|1_{y_{2} = 0}\left|\frac{1}{|x_{0} - y|} - \frac{1}{|x - y|}\right|\, dy$ is small, but this is bounded above by $\int_{\mathbb{R}^2}|f(y)|\left|\frac{1}{|x_{0} - y|} - \frac{1}{|x - y|}\right|\, dy$ and now we proceed as in the proof of showing $(1/|x|) \ast f$ is continuous.
Aug
6
comment Where is $\int_{-\infty}^{\infty}\frac{f(y)}{\sqrt{(y - x_{1})^{2} + x_{2}^{2}}}\, dy$ continuous?
I've been trying to base the proof of the proof that $(1/|x|) \ast f$ is continuous, however, the issue is that now the $1/\sqrt{(y - x_{1})^{2} + x_{2}^{2}}$ part is not a convolution so I can't easily do a change of variables to put the translation onto $f$. I was wondering what else I could try to do.
Aug
6
asked Where is $\int_{-\infty}^{\infty}\frac{f(y)}{\sqrt{(y - x_{1})^{2} + x_{2}^{2}}}\, dy$ continuous?
Jul
8
asked Bound on the size of the solution to $u_{t} - \Delta u - u \leq 0$
May
23
accepted $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@user26857: Implicitly throughout, we're viewing $f$ and $g$ now as functions over $\mathbb{F}_{p}$. So we have $f'(x)g(x) = f(x)g'(x)$ for all $x \in \mathbb{F}_{p}$. If one really wanted to be pedantic, I suppose one could add the phrase "over $\mathbb{F}_{p}$" everywhere, but I definitely agree that Matt's solution is correct.
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@barto: It turns out that that's not how I define $\deg f$. I define it as $\max(\deg g, \deg h)$, sorry for the confusion.
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@user26857: It turns out I was misinterpreting what I was reading, I have changed to question to better reflect what I want to ask.
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
rewrote the whole question to be more clear
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
added 58 characters in body
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
added 58 characters in body
May
23
asked $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
May
3
accepted Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
May
3
comment Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
$N$ is fixed and large.
May
3
revised Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
clarified role of n
May
3
asked Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
Dec
21
awarded  Popular Question