Prove that $L_{2}^{D}$ is an inner product space.

Question

I want to prove the following:

What is the inherited inner product from $l_{2}$, how can I prove that it is an inner product space? could anyone help me please?

Answer 1

Let $(x_n)_{n=1}^\infty$, $(y_n)_{n=1}^\infty$ be two sequences in $\ell^D_2$ such that $x_n = 0$ for $n \in K$ and $y_n = 0$ for $n \in L$, where $K, L \subseteq \mathbb{N}$ are two sets with finite complements.

Then we have $x_n + y_n = 0$ for $n \in K \cap L$, whose complement is finite so $(x_n)_{n=1}^\infty + (y_n)_{n=1}^\infty = (x_n + y_n)_{n=1}^\infty \in \ell^D_2$.

Similarly, for $\lambda \ne 0$ we have $\lambda x_n = 0$ for $n \in K$ so $\lambda(x_n)_{n=1}^\infty = (\lambda x_n)_{n=1}^\infty \in \ell^D_2$. For $\lambda = 0$, obviously we have $\lambda(x_n)_{n=1}^\infty = 0 \in \ell^D_2$.

Thus, $\ell^D_2$ is a vector subspace of $\ell_2$.

Now, let $\langle\cdot, \cdot\rangle$ be the inner product on $\ell_2$. To show that its restriction is an inner product on $\ell^D_2$, we have to show:

$$\langle x,x\rangle \ge 0, \quad\forall x \in \ell^D_2$$ $$\langle x,x\rangle = 0 \implies x = 0, \quad\forall x \in \ell^D_2$$ $$\langle x,y\rangle = \overline{\langle y,x\rangle}, \quad\forall x,y \in \ell^D_2$$ $$\langle \lambda x + \mu y,z\rangle = \lambda \langle x,z\rangle + \mu \langle y,z\rangle, \quad\forall x,y,z \in \ell^D_2, \forall \lambda, \mu \in \mathbb{C}$$

But every single of these properties already holds for all vectors in $\ell_2$, so it certainly holds for all vectors in $\ell^D_2 \subseteq \ell_2$.

Therefore, $\langle \cdot, \cdot\rangle\Big|_{\ell^D_2}$ is an inner product on $\ell^D_2$.