How to construct such a Linearly Independent subset of V having n elements

Solution 1:

This is the dual space of an infinite dimensional space, so it's definitely infinite dimensional. However, you want to show this by explicitly showing an infinite linearly independent set.

An easy way to define functionals on $C^\infty(\mathbb{R})$ is to take evaluations at points: for $x\in\mathbb{R}$, define $\hat{x}\colon C^\infty(\mathbb{R})\to\mathbb{R}$ by $\hat{x}(f)=f(x)$.

Can we say that if $x_1,\dots,x_n$ are distinct points then $\{\hat{x}_1,\dots,\hat{x}_n\}$ is linearly independent? Suppose we have a linear combination $$ a_1\hat{x}_1+a_2\hat{x}_2+\dots+a_n\hat{x}_n=0 $$ Thus, for every $f\in C^\infty(\mathbb{R})$, we have $$ a_1f(x_1)+a_2f(x_2)+\dots+a_nf(x_n)=0 $$ Now it's easy. Take a polynomial $f$ such that $f(x_1)=1$ and $f(x_i)=0$ for $i\ne 1$. This proves that $a_1=0$. Finish up.