Inner product of a finite sum

Question

I have been working on this question for very long, but I can't find the right answer. I know that I have to check the properties for an inner product, but it gives me a hard time. The question is:

Can anyone help me?

Answer 1

(a) Let $f,g,h\in S$ and $c\in\mathbb{R}$, and let $M,N\in\mathbb{N}$ be such that $f(n_1)=g(n_2)=0$ for all $n_1>M$ and $n_2>N$. To verify $\langle\cdot,\cdot\rangle$ is an inner product, we have to check the followings.

• We have \begin{align} \langle f,g\rangle =\sum_{n=0}^\infty f(n)g(n) =\sum_{n=0}^N f(n)g(n) =\sum_{n=0}^M g(n)f(n) =\sum_{n=0}^\infty g(n)f(n) =\langle g,f\rangle. \end{align}

• We have \begin{align} \langle f+h,g\rangle &=\sum_{n=0}^\infty (f+h)(n)g(n) =\sum_{n=0}^{N} (f+h)(n)g(n) =\sum_{n=0}^{N} [f(n)+h(n)]g(n)\\ &=\sum_{n=0}^{N} [f(n)g(n)+h(n)g(n)] =\sum_{n=0}^{N} f(n)g(n)+\sum_{n=0}^{N}h(n)g(n)\\ &=\sum_{n=0}^{\infty} f(n)g(n)+\sum_{n=0}^{\infty}h(n)g(n) =\langle f,g\rangle+\langle h,g\rangle. \end{align}

• We have \begin{align} \langle cf,g\rangle &=\sum_{n=0}^\infty (cf)(n)g(n) =\sum_{n=0}^{N} (cf)(n)g(n) =c\sum_{n=0}^{N} f(n)g(n) =c\sum_{n=0}^{\infty} f(n)g(n) =c\langle f,g\rangle. \end{align}

• We have \begin{align} \langle f,f\rangle &=\sum_{n=0}^\infty f(n)f(n) =\sum_{n=0}^{M} [f(n)]^2 \ge 0. \end{align}

(b) We first show that $B$ is a basis for $S$. Given distinct vectors $e_{n_1},e_{n_2},\ldots,e_{n_m}\in B$ and scalars $a_1,a_2,\ldots,a_m\in\mathbb{R}$ such that $\displaystyle\sum_{i=1}^ma_ie_{n_i}={\it 0}$. Then for $j=1,2,\ldots,m$, $$0=\left(\sum_{i=1}^ma_ie_{n_i}\right)(n_j)=\sum_{i=1}^ma_ie_{n_i}(n_j) =\sum_{i=1}^ma_i\delta_{n_in_j}=a_j,$$ where $\delta_{n_in_j}$ is the Kronecker delta function. Hence $B$ is linearly independent. Next, given $f\in S$, choose $N\in\mathbb{N}$ such that $f(n)=0$ for all $n>N$. Define $f^\ast=\displaystyle\sum_{n=1}^Nf(n)e_n$. For $k>N$, we have $f^\ast(k)=0=f(k)$. For $k\le N$, $$f^\ast(k)=\left(\sum_{n=1}^Nf(n)e_n\right)(k) =\sum_{n=1}^Nf(n)e_n(k)=\sum_{n=1}^Nf(n)\delta_{nk}=f(k).$$ So $f^\ast=f$, which shows that $f$ is a linear combination of vectors in $B$. Hence $B$ is a basis for $S$. Finally, we show that $B$ is orthonormal. For each $m,n\in\mathbb{N}$ with $n\ne m$, we have $$\langle e_m,e_n\rangle =\sum_{k=1}^\infty e_m(k)e_n(k) =\sum_{k=1}^\infty\delta_{mk}\delta_{nk} =\delta_{mn}\delta_{nm}+\delta_{mn}\delta_{nn}=0+0=0. $$ Also, $\Vert e_n\Vert^2=\langle e_n,e_n\rangle =\displaystyle\sum_{k=1}^\infty [e_n(k)]^2 =\sum_{k=1}^\infty \delta_{nk}^2=\delta_{nn}^2=1$. Hence $B$ is orthonormal.