Banach space it isn't Hilbert space

Question

How can give me two or three examples about Banach spaces which it is not Hilbert spaces with proof ( I mean why it isn't Hilbert spaces ) ?

Answer 1

First of all you should understand that what is meant when one says that "$\textit{Not all normed linear spaces are inner product spaces.}$" Here is the explanation.

An $\textit{inner product space}$ is a vector space with an inner product defined on it. An inner product on $X$ defines a norm on $X$ given by \begin{equation} \|x\| = \sqrt{\langle x,x \rangle} \end{equation} and a metric on $X$ given by \begin{equation} d(x,y) = \|x-y\| = \sqrt{\langle x-y,x-y\rangle}. \end{equation} A complete inner product space is called a $\textit{Hilbert space}$. Hence inner product spaces are normed linear spaces and Hilbert spaces are Banach spaces.

A simple straightforward calculation shows that a norm on an inner product space satisfies the important $\textit{parallelogram law}$ \begin{equation} \|x+y\|^2 +\|x-y\|^2 = 2(\|x\|^2+\|y\|^2). \end{equation} We conclude that if a norm does not satisfy the parallelogram law, it cannot be obtained from an inner product. Without risking misunderstanding we may thus say:

$\textit{Not all normed linear spaces are inner product spaces.}$

$\textit{The space $l_p$ with $p\neq 2$ is not an inner product space, hence not a Hilbert space.}$

Our statement means that the norm of $l_p$ with $p \neq 2$ cannot be obtained from an inner product. We prove this by showing that the norm does not satisfy the parallelogram law. In fact, let us take $x=(1,1,0,0,\ldots) \in l_p$ and $y = (1,-1,0,0,\ldots) \in l_p$. Then \begin{align*} \|x\|=\|y\|=2^{\frac{1}{p}}, & & \|x+y\|=\|x-y\|=2. \end{align*} We now see that $\textit{parallelogram law}$ is not satisfied if $p\neq 2$. We also know that $l_p$ is complete. Hence $l_p$ with $p\neq 2$ is a Banach space which is not a Hilbert space.

Answer 2

The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work.

One way to prove that this norm does not come from any inner product is to prove that it does not satisfy the parallelogram law, which says $$ \|x+y\|^2+\|x-y\|^2 = 2\|x\|^2+2\|y\|^2. $$ The parallelogram law holds in inner product spaces.