How to prove that $L^p [0,1]$ isn't induced by an inner product? for $p\neq 2$

Question

I'd like to know how could I prove that $L^p [0,1]$ isn't induced by an inner product? (For $p\neq 2$, including $p=\inf$).

It is clear to me that I would need to find two functions $f$, $g$ in $L^p$ that would not fulfill the parallelogram-law.

I'm having trouble finding these functions (for both finite and infinite cases). Could you guide me how to find such functions? what intuition is there for these kind of questions? What would be special about them?

Any guidenss would be blessed. Thank you

Answer 1

I'll give you a hint.

Let $$f(x)=\begin{cases}2,x\in[0,3/4],\\0,x>3/4,\end{cases}\quad g(x)=f(1-x).$$

Then the parallelogram law says that in hilbert spaces we have $$2\|f\|^2+2\|g\|^2=\|f-g\|^2+\|f+g\|^2.$$

Can you calculate the norms above?

Answer 2

How to find these...

By linearity of the integral one has: $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|^p\mathrm{d}\lambda=\int|f|^p\mathrm{d}\lambda+\int|g|^p\mathrm{d}\lambda$$

So take a block and shift it slightly: $$f:=\chi_{[0,1]}:\quad f_\varepsilon(x):=f(x-\varepsilon)$$

Then for an appropriate choice: $$\|f+f_\varepsilon\|_p^2+\|f-f_\varepsilon\|_p^2\neq2\|f\|_p^2+2\|f_\varepsilon\|_p^2$$

...besides, this idea should work for any nontrivial function not only blocks.

Answer 3

For the case in which $1 \leq p < \infty$:

Let $\displaystyle f = \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A$ and $\displaystyle g = \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B$ where $A, B$ both have nonzero finite measure, are disjoint and $\chi$ is the indicator function.

Notice that $$\|f\|_p = \bigg( \int \vert f \vert^p \, d\mu \bigg)^{1/p} = \bigg( \int \frac{1}{\mu(A)} \chi_A \, d\mu \bigg)^{1/p} = \bigg( \int_A \frac{1}{\mu(A)} \, d\mu \bigg)^{1/p} = 1$$ and similarly $\|g\|_p = 1$.

Observe that

$$\begin{align*} \|f + g\|_p&= \bigg( \int \vert f + g \vert^p \, d\mu \bigg)^{1/p} \\ &=\bigg( \int \bigg \vert \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A + \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B \bigg \vert^p \, d\mu \bigg)^{1/p} \\ &= \bigg( \int \frac{1}{\mu(A)} \chi_A + \frac{1}{\mu(B)} \chi_B \, d\mu \bigg)^{1/p} \\ &= \bigg( \int_A \frac{1}{\mu(A)} \, d\mu + \int_B \frac{1}{\mu(B)} \, d\mu \bigg)^{1/p} \\ &= 2^{1/p} \end{align*}$$

Note: We use the fact that $A$ and $B$ are disjoint and get $\chi_A \cdot \chi_B = 0$ when moving from line 2 to line 3 above.

Similar reasoning shows $\|f - g\|_p = 2^{1/p}$.

Recall that in an inner product space, the Parallelogram law holds: $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \|f\|_p^2 + 2 \|g\|_p^2$.

But $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \cdot 2^{2/p}$ and $2 \|f\|_p^2 + 2 \|g\|_p^2 = 4$ which means that we must have $2^{2/p} = 2$ which only happens if $p = 2$.

For the case in which $p = \infty$:

Let $f = \chi_A$ and $g = \chi_B$ where $A, B$ are disjoint and both have nonzero measure.

Notice that $\|f\|_\infty = \|g\|_\infty = 1$ and $\|f + g\|_\infty = \|f - g\|_\infty = 1$. The Parallelogram Law doesn't hold because $\|f + g\|_\infty^2 + \|f - g\|_\infty^2 = 2$ but $ 2 \|f\|_\infty^2 + 2 \|g\|_\infty^2 = 4$