Define an element of $B$ such that $f(x)=0 $ for $ x<0$ and $f(x)>0$ for $x<0$.

71 Views Asked by At

I came across a question sheet that has no solutions and I'm particularly stuck on these 3.

Let $B$ denote collection of functions from $\mathbb{R}^n $ to $ \mathbb{R} $ that are $p$ times differentiable for every natural number $p$.

$B$ is a vector space under addition and scalar multiplication operations.

  • For $n=1$, how would I define element of $B$ such that $f(x)=0 $ for $ x<0$ and $f(x)>0$ for $x<0$ ?

  • For $f_1,...,f_n \in B $, how would I prove $D=f_1\partial_1 + ... +f_n\partial_n $defines linear map from $B$ to $B$ such that $D(ab)=aD(b)+D(a)b $ ?

  • What is an $n$ dimensional space $K$ of derivations such that $[k,k']=0$ for $k,k' \in K$ ?

Sorry for the length. If anyone could offer step by step solutions for any of these I would really appreciate it, I have vague ideas but I think they're incorrect.

1

There are 1 best solutions below

2
On

One

One possible function, for your first question, is $$f(x)=\begin{cases}0& x\le 0 \\ x^{p+1} & x>0\end{cases}$$

Two

$\partial_i$ is a linear map. That is, $$\partial_i(\alpha f+\beta g)=\alpha \partial_i u+\beta\partial_i v, \forall \alpha,\beta\in\mathbb{R}, \forall u,v\in B.$$ Now,

\begin{align}D(\alpha u+\beta v)& \\ &= (\sum_i f_i\partial_i)(\alpha u+\beta v)\\&=\sum_i (f_i\partial_i)(\alpha u+\beta v)\\&=\sum_i f_i\partial_i(\alpha u+\beta v)\\&=\sum_i f_i(\alpha\partial_i u+\beta\partial_i v)\\&=\alpha\sum_i f_i\partial_i u+\beta\sum_if_i\partial_i v)\\&=\alpha D(u)+\beta D(v),\end{align} which shows that $D$ is linear.

Since $\partial_i(uv)=v\partial_iu+u\partial_iv$ we can prove as above that $D(uv)=vDu+uDv$

Three

We have that $\partial_i$ is a derivation for all $i.$ Moreover note that $\{\partial_1,\cdots,\partial_n\}$ is the basis of an $n$-dimensional space. How to show that they are linearly independent? Note that $$(\sum_i\partial_i)=0\implies (\sum_i\partial_i)x_j=0 \implies a_j=0,\forall j.$$

Finally, if $p\ge 2$ then we have $$[\partial_i,\partial_j]f=\partial_i\partial_jf-\partial_j\partial_if=0.$$