# $L^2$ is the only Hilbert space , Parallelogram Law and particular $f(t),g(t)$

Let the space $$C([0,1])$$ and consider the norm $$\forall p \in \mathbb{N}$$ $$\forall f \in C([0,1]), ||f ||_{L^P}= \left ( \int^1_0 |f(t)|^P dt \right )^{\frac{1}{p}}$$

knowing that when $$p=2$$, this norm is the norm induced by the innder product

$$\forall f,g \in C([0,1]), =\int^1_0 f(t)\overline{g(t)} dx$$

The goal of this excersice is to prove that if $$p \neq 2$$ $$||.||_p$$ is not a norm induced by an inner product (i.e $$L^2$$ is the only hilbert space among all $$L^p$$ spaces. )

To do so, study when the parallelogram law holds. Hint: consider the functions

$$f(t)=\frac{1}{2}-t; g(t)= \begin{matrix} \frac{1}{2} -t & \text{ if } 0 \leq t \leq \frac{1}{2} \\ t-\frac{1}{2} & \text{ if } \frac{1}{2}

Parallelogram Law $$M=C([0,1])$$

Let $$(M,<.,.>)$$ is an inner product space, where induced norm $$|| .|| =\sqrt{<.,.>}$$ then $$\forall x,y \in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2)$$

Using u-sub $$u=(1/2 -t)$$ and $$\frac{du}{dt}=-1$$ so $$dt=-du$$

\begin{aligned} ||f|| &= \int^1_0 (|f(t)|^p dt)^{1/p}= \int^1_0 (|1/2 -t|^p dt)^{1/p} \\ &=\int^{u(1)}_{u(0)} (u^p -du)^{1/p} \\& =(\int^{-1/2}_{1/2} u^p du)^{1/p} \\ &= \left [ \left (\frac{u^{p+1}}{p+1} \right )^{1/p} \right]^{-1/2}_{1/2} \\&= \left [\frac{u^{1+1/p}}{(p+1)^{1/p}} \right]^{-1/2}_{1/2} \\&=\frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}} \end{aligned}

working on $$||f-g ||,||f+g||,||g||$$

Guessing big picture that the parallelogram law only works when $$p=2$$ so it is only induced norm when $$p=2$$

• When $p=2$ inner product does induce the norm, and when $p\neq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products – user160738 Nov 8 '16 at 19:49

Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<\infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $p\ne q$ these are not isomorphic.
• How do you prove that $L^p$ and $L^q$ are not isomorphic for $p \ne q$? – gerw Nov 8 '16 at 19:59
• This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one.. – gerw Nov 8 '16 at 20:05