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

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

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

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

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 $p=2$

  • $\begingroup$ 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 $\endgroup$
    – 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.

  • $\begingroup$ How do you prove that $L^p$ and $L^q$ are not isomorphic for $p \ne q$? $\endgroup$
    – gerw
    Nov 8 '16 at 19:59
  • 1
    $\begingroup$ This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one.. $\endgroup$
    – gerw
    Nov 8 '16 at 20:05

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