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1) Define : $\langle z\rangle := (1+|z|^2 ) ^\frac{1}{2} $ for all $z \in \mathbb{C} $. Prove : $\langle x+y\rangle \leq 2\langle x\rangle\langle y\rangle $ for all $x,y \in \mathbb{R} ^N$ .

2) Define: $ S^\beta := \left\{ f: \mathbb{R} \to \mathbb{C} : f \in C^\infty(\mathbb{R}), \ \left|f^{(n)}(x)\right|:= \left|\frac{d^n f}{dx^n}\right| \leq c_n \langle x\rangle^{\beta -n} \right\} $ for some $c_n<\infty$, all $x \in \mathbb{R} $ and all $ 0\leq n \in \mathbb{Z}$. Then, define: $ \mathbb{A} := \cup_{\beta<0 } S^\beta $. Prove $\mathbb{A}$ is an algebra under pointwise multiplication.

I'm pretty much clueless regarding (1). As for 2 - I can't understand which notion of an "algebra" they mean... If it's an algebra like in "sigma-algebra", then why do we need the pointwise multiplication? can you help me prove this?


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#1 might be easier to prove if you square both sides first. – Antonio Vargas May 11 '12 at 20:51
You define brackets for complex numbers and then apply to vectors of $\mathbb{R}^N$. I don't understand... – Norbert May 11 '12 at 20:54
By the way, the correct notion of an algebra in this context, is the one described here: (this has nothing to do with the notion of a $\sigma$-algebra). – Martin Wanvik May 11 '12 at 21:00
For (1), expand and use the resulting identity $(|x|-|y|)^2 \geq 0$. – copper.hat May 11 '12 at 21:31
up vote 1 down vote accepted

I assume that $x,y\in\mathbb{C}$ since that is where $\langle\cdot\rangle$ is defined. Here are some hints:

1) Let $x=a+ib$ and $y=c+id$. Then $$ \begin{align} \langle x+y\rangle^2 &=1+(a+c)^2+(b+d)^2\\ &=1+a^2+b^2+c^2+d^2+2ac+2bd\\ &\le1+a^2+b^2+c^2+d^2+(a^2+c^2)+(b^2+d^2)\tag{1} \end{align} $$ and $$ \begin{align} \langle x\rangle^2\langle y\rangle^2 &=(1+a^2+b^2)(1+c^2+d^2)\\ &=1+a^2+b^2+c^2+d^2+(a^2c^2+a^2d^2+b^2c^2+b^2d^2)\\ &\ge1+a^2+b^2+c^2+d^2\tag{2} \end{align} $$ Compare $(1)$ and $(2)$.

2) You are supposed to show that if $f,g\in S^\beta$, then $fg\in S^\beta$. Think product rule.

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Thanks! !!!!!!! – joshua May 12 '12 at 7:30
Seeing as you got accepted I think you should also get an upvote. – Rudy the Reindeer May 18 '12 at 21:10

1) Upon squaring, what you need to show is $$ 1+|x|^2+|y|^2+2|x||y|=1+|x+y|^2 \leq 4(1+|x|^2)(1+|y|^2) = 4 + 4|x|^2+4|y|^2+4|x|^2|y|^2, $$ which reduces the problem to showing $$ 3+3(a^2+b^2)+4a^2b^2\geq2ab, $$ for all $a,b\geq0$.

2) Here what you need to show is that the product $fg\in\mathbb{A}$ for any $f,g\in\mathbb{A}$.

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Thanks! !!!!!!! – joshua May 12 '12 at 7:30

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