In an inner product space, the inner product determines the length of a vector. More generally, we can specify the length directly by defining a norm.
- Positive definiteness. \(\|\mathbf{v}\|\ge 0\), with equality if and only if \(\mathbf{v}=\mathbf{0}\).
- Absolute homogeneity. \(\|\alpha\mathbf{v}\|=|\alpha|\,\|\mathbf{v}\|\).
- Triangle inequality. \(\|\mathbf{u}+\mathbf{v}\|\le\|\mathbf{u}\|+\|\mathbf{v}\|\).
Every norm induces a metric on \(V\) given by
Example 1
Let \(C[a,b]\) be the space of continuous functions \(f\colon[a,b]\to\mathbb{F}\). Its supremum norm is
Every Cauchy sequence in this norm converges uniformly to a function on \([a,b]\). The uniform limit of continuous functions is continuous, so the limit belongs to \(C[a,b]\). Thus \(C[a,b]\) is a Banach space.
Example 2
Let \((H,\langle\cdot,\cdot\rangle)\) be a Hilbert space. Its inner product induces the norm
A Hilbert space is complete in this norm by definition. Hence every Hilbert space is a Banach space.
Example 3
Let \(\mathbb{T}=\mathbb{R}/(2\pi\mathbb{Z})\) denote the circle. The space \(L^1(\mathbb{T})\) consists of equivalence classes of measurable functions \(f\colon\mathbb{T}\to\mathbb{C}\) satisfying
Two functions represent the same element if they agree almost everywhere. With the normalized norm
\(L^1(\mathbb{T})\) is a Banach space. Its norm fails the parallelogram identity, so no inner product induces this norm.
Example 4
The space \(\ell^1(\mathbb{N})\) consists of all absolutely summable sequences. It is given by
It is a Banach space with the norm
Example 5
The space \(c_0\) consists of all sequences that converge to zero. It is given by
Every sequence that converges to zero is bounded, so its supremum norm is finite. With the norm
\(c_0\) is a Banach space. Indeed, \(c_0\) is a closed subspace of the Banach space of bounded sequences in the next example.
Example 6
The space of bounded sequences is
Its norm is
To see why this space is complete, let \(x^{(k)}=(x_n^{(k)})_{n=1}^{\infty}\) be Cauchy in the supremum norm. For each fixed \(n\), the scalars \(x_n^{(k)}\) are Cauchy and have a limit \(x_n\). Given \(\epsilon>0\), choose \(K\) so that \(\|x^{(k)}-x^{(j)}\|_\infty \lt \epsilon/2\) for all \(k,j\ge K\). Letting \(j\to\infty\) in each coordinate gives
Taking \(k=K\) shows that \(x=(x_n)\) is bounded, since \(x^{(K)}\) is bounded. Taking the supremum over \(n\) shows that \(\|x^{(k)}-x\|_\infty\to 0\). Thus \(\ell^\infty(\mathbb{N})\) is a Banach space.
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi