Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

3

Take $N(x,y) = |x|$ on $\mathbb{R}^2$.

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A Hamel basis is defined as a maximal linearly independent set. Every vector space has a Hamel basis and this well known result is proved using Zorn's Lemma. In your example $B$ is not a Hamel basis. (And you have proved that it is not one! You know that there is an element $x$ which is not a finite linear combination of members of $B$. This implies $B \... 2 If your metric space contains an isolated point$p$and at least one other point, a unit mass at$p$can't be approximated by continuous measures. 1 There’s the trivial seminorm:$N(x)=0$for all$x\in V$. 1 Since${\rm sech}=1/\cosh$is an even function we can replace$f$with$|f|$. Assume for starters that$|f|$is continuous; that$|f(x)|$goes fast enough to$\infty$for$x\to \infty$such that OP's integral $$I(b)~:=~b\int_{\mathbb{R}_+}\! \mathrm{d}x~{\rm sech}^2(b|f(x)|), \qquad b>0,\tag{1}$$ is convergent; that$x=c>0$is the only zero for$|...

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For $|x|>1$ let $p_n(x)=0.$ For $|x|\le 1$ let $p_n=c_n(1-x^2)^n$ where $c_n\cdot\int_{-1}^1(1-x^2)^ndx=1.$ By 14.8.2.(b) we have $c_n\le \sqrt n\,.$ The idea is to show that $p_n\to 0$ uniformly on $[-1,-\delta]\cup [\delta,1]$ for any fixed $\delta \in (0,1).$ So, given any $\epsilon^*$ in $(0,1)$ we find some "small" $x_n$ in $(0,1)$ such that $x_n\... 1 After using the trigonomy formulas for$(\sin x)^n$and applying partial integration, I've got with$n\in\mathbb{N}$: $$J(n)=\int\limits_0^\pi\left(\frac{\sin x}{x}\right)^n dx = \frac{1}{(n-1)!}\sum\limits_{k=0}^{\lfloor (n-1)/2\rfloor}(-1)^k{\binom n k}\left(\frac{n}{2}-k\right)^{n-1}\text{Si}((n-2k)\pi)$$ 1 As stated, the assertion is not true. Take$X=[-1,1]$and $$\mathcal A=\{\lambda 1 + f:\ f|_{[0,1]}=0\}.$$ Then$\mathcal A$satisfies your conditions, and it contains no polynomials other than the constants. Edit: with access to the link, it looks like you have misunderstood a few things. What happens is this: the algebra$\mathcal A\$ is closed; ...

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