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seen Oct 8 at 8:44

Oct
18
comment The product of a uniformly continuous function and a bounded continuous function is uniformly continuous
You may start with $g\equiv 1$.
Oct
18
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
@EricAuld: I don't know, but if you are familiar with Jensen's inequality, then you can apply it to different convex functions to produce kinds of inequalities like the one you mentioned as you wish.
Oct
18
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
@EricAuld: You are welcome!
Oct
18
answered If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational
Oct
18
comment Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
@Sanchez: You are welcome and thank you for your help in advance!
Oct
17
revised Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
deleted 99 characters in body
Oct
17
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
The flaw is you didn't obtain any control of the integral of $\frac{1}{|f(\sigma)|}$ over $S^{n-1}$.
Oct
17
answered Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
Oct
17
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
By Fubini's theorem, the integral related to the GM will diverge when $\alpha_i\le n$ for some $i$.
Oct
17
reviewed Approve complete metric on a Riemann Surface
Oct
17
comment Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
Hi, Sanchez. Since the question is closely related to $T_n$ and since $T_n$ satisfies a recurrence relation $T_n(x)=2xT_{n-1}(x)-T_{n-2}(x)$, I wonder if you have considered the following statement. If $f$ is a polynomial with $\deg f=n\ge 2$, then there exist polynomials $g$ and $h$, such that $f(x)=2xg(x)-h(x)$, $\deg g\le n-1$, $\deg h\le n-2$ and $\|g\|,\|h\|\le \|f\|$. Here $\|f\|$ denotes the maximum modulus of $f$ on $[-1,1]$. I have no clue whether statement is true or not. Sorry for bothering.
Oct
16
revised Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$
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Oct
16
answered Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$
Oct
15
comment An abstract definition of the cotangent space to a smooth manifold.
What about GTM 94, Foundations of Differentiable Manifolds and Lie Groups, by Frank W. Warner?
Oct
11
answered Functions satisfying $(b-a)f'(\tfrac{a+b}{2}) = f(b)- f(a)$
Oct
8
comment If $f(x)$ is positive and decreasing, can $xf(x)$ have more than one maxima?
An alternative example: $f(x)=1-x$ when $x\in[0,\frac{1}{3}]\cup [\frac{2}{3},1]$; $f(x)=\frac{2}{9x}$ when $x\in[\frac{1}{3},\frac{2}{3}]$.
Oct
8
comment Question about convolution
What if $f\equiv 1$ and $g$ is not integrable?
Oct
7
comment How prove this $f(a)\le f(b)$
@nanchangjian: Once again, you are welcome.
Oct
7
comment How prove this $f(a)\le f(b)$
@nanchangjian: You are welcome. By definition, $x_y$ is at least a lower bound of the set $E=\{x\in[a,b]: f_\epsilon(x)<y\}$, which means that if $x<x_y$, $f(x)\notin E$, i.e. $f(x)\ge y$. Then by continuity, $f(x_y)\ge y$(actually, since $x_y$ is the greatest lower bound, $f(x_y)= y$).
Oct
7
answered How prove this $f(a)\le f(b)$