Solution. Write \((x,y)=\sum _{i=1}^{n}x_i\overline {y_i}\) and check the axioms.

Linearity in the first argument

For scalars \(\alpha ,\beta \), \[(\alpha x+\beta y,z)=\sum _{i=1}^{n}(\alpha x_i+\beta y_i)\overline {z_i} =\alpha \sum _{i=1}^{n}x_i\overline {z_i}+\beta \sum _{i=1}^{n}y_i\overline {z_i} =\alpha (x,z)+\beta (y,z).\]

Conjugate symmetry

\[\overline {(y,x)}=\overline {\sum _{i=1}^{n}y_i\overline {x_i}} =\sum _{i=1}^{n}\overline {y_i}x_i=(x,y).\]

Positive definiteness

\[(x,x)=\sum _{i=1}^{n}x_i\overline {x_i}=\sum _{i=1}^{n}|x_i|^2\geq 0 ,\] and this vanishes only if every \(|x_i|=0\), that is \(x=0\). Note that \((x,x)\) is automatically real, as conjugate symmetry requires.

Conjugate symmetry together with linearity in the first slot gives conjugate linearity in the second, so the form is sesquilinear rather than bilinear — the price of working over \(\mathbb {C}\) instead of \(\mathbb {R}\).

Solution.

The series converges

This needs checking before anything else. For each \(N\), the Cauchy–Schwarz inequality for finite sums gives \[\sum _{n=1}^{N}|x_n\overline {y_n}| \leq \left (\sum _{n=1}^{N}|x_n|^2\right )^{\frac 12} \left (\sum _{n=1}^{N}|y_n|^2\right )^{\frac 12}\leq ||x||\,||y|| .\] The partial sums of \(\sum |x_n\overline {y_n}|\) are therefore bounded, so the series converges absolutely and \((x,y)\) is a well-defined complex number.

The axioms

Each is inherited from the finite case by letting \(N\rightarrow \infty \), the limits being legitimate because all the series converge. Linearity in the first argument and conjugate symmetry follow termwise as in the previous example, and \[(x,x)=\sum _{n=1}^{\infty }|x_n|^2=||x||^2\geq 0 ,\] which vanishes only when every \(x_n=0\).

The restriction to \(p=2\) is essential: for \(p\neq 2\) the series need not even converge, and the norm of \(l_p\) fails the parallelogram law, so no inner product can induce it.