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 Apr 27 asked If $f$ is Schwartz, does there exist a positive Schwartz function $g$ such that $|f(x)| \leq g(x)$? Jan 31 asked Showing $y''' + y'^2 = 0$ with $y(0) = 0$, $y'(0) = 1$, $y''(0) = 0$ is asymptotic to $x$ rigorously Jan 15 awarded Yearling Dec 4 comment Does $\sum_{n = 1}^{\infty}(\sup_{x \in [0, 1]}f_{n}(x)) < \infty$ imply $\sum_{n = 1}^{\infty}f_{n}(x)$ converges uniformly? @neth: If $\sum_{n = 1}^{\infty}(\sup_{x \in [0, 1]}|f_{n}(x)| < \infty$, then does $\sum_{n = 1}^{\infty}f_{n}(x)$ converges uniformly by the Weierstrass M-test? Dec 4 asked Does $\sum_{n = 1}^{\infty}(\sup_{x \in [0, 1]}f_{n}(x)) < \infty$ imply $\sum_{n = 1}^{\infty}f_{n}(x)$ converges uniformly? Dec 4 accepted Explicit Schwartz $f$ such that $f = 1$ in $B(0, 1)$ and $\widehat{f} = 0$ outside $B(0, 1)$ Nov 24 awarded Popular Question Oct 22 comment Explicit Schwartz $f$ such that $f = 1$ in $B(0, 1)$ and $\widehat{f} = 0$ outside $B(0, 1)$ Why would then $f$ be the restriction to $\mathbb{R}^{n}$ of an entire function in $\mathbb{C}^{n}$? What if instead I require that $|f| \geq 1$ on the ball of radius 1? Oct 22 asked Explicit Schwartz $f$ such that $f = 1$ in $B(0, 1)$ and $\widehat{f} = 0$ outside $B(0, 1)$ 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.