# Parallelogram law in $L_1$ space

Exercise 5.5 from Capinski's and Kopp's book "Measure, Integral and Probability" asks to show that it is impossible to define an inner product on the space $L^1([0,1])$. In order to get this result we need to show that parallelogram law

$$\Vert f+g \Vert^2 + \Vert f-g \Vert^2 = 2( \Vert f\Vert^2 + \Vert g \Vert^2 )$$

does not hold for $\Vert \cdot \Vert_1$ where $$\Vert f \Vert_1=\int_0^1|f(x)|\mathsf dx.$$ The hint in the book suggests to consider two following functions:

\begin{align} f(x)&=\frac12-x\\ g(x)&=x-\frac12. \end{align}

However, $g(x)=-f(x)$ and thus $\Vert g \Vert_1 = \Vert f \Vert_1$ . Moreover, $\Vert f+g \Vert_1 = 0$ and $$\Vert f-g \Vert_1 = \Vert 2f \Vert_1 = 2\Vert f \Vert_1.$$

Finally I get:

\begin{align} \Vert f+g \Vert_1^2 + \Vert f-g \Vert_1^2 &= 4\Vert f \Vert_1^2\\ &= 2(2\Vert f \Vert_1^2)\\&= 2( \Vert f\Vert_1^2 + \Vert g \Vert_1^2 ) \end{align}

which in turn shows that actually Parallelogram law holds.

What did I miss?

May it be related to the fact that $g$ is a scaled version of $f$?

## 2 Answers

There is nothing wrong about your argumentation.

Here is a "real" counterexample: Set

$$f(x) := 1_{[0,1/2]}(x) \qquad \text{and} \qquad g(x) := 1_{[1/2,1]}(x).$$

Then

$$\|f\|_1 = \|g\|_1 = \frac{1}{2}$$

and therefore

$$2 (\|f\|_1^2+ \|g\|_1^2) = 1.$$

On the other hand, it is not difficult to see that $$\|f+g\|_1 = \|f-g\|_1 = 1.$$

Hence,

$$\|f+g\|_1^2 + \|f-g\|_1^2 = 2 \neq 1 = 2 (\|f\|_1^2+ \|g\|_1^2).$$

• thanks. It is obvious that that parallelogram law holds for your $f$ and $g$. Could you comment on the case when $g=-f$, because I thought that the law should NOT hold for all functions in $L^1([0,1])$. – Ruth90 Jul 30 '15 at 15:28
• @Ruth90 Actually, the parallelogram law does not hold for the $f$ and $g$ from my answer. You wanted to find $f$,$g$ such that the parallelogram law doesn't hold, right? If $g=-f$, then the parallelogram law does hold true; that's exactly what you proved. – saz Jul 30 '15 at 15:31
• sorry you are right - in your case the law does NOT hold, which is fine since, if I am correct, the law holds just in $L^2$ space (I am working with $L^p$ spaces). My question would be why does parallelogram does hold in the case of $g=-f$? Which in turn means that the $\Vert ^. \Vert_1$ can be induced by inner product. – Ruth90 Jul 30 '15 at 15:35
• @Ruth90 Well, why not? That the parallelogram law doesn't hold true in $L^1$ doesn't mean that the identity fails for all $f,g \in L^1$; it simply means that there are $f,g$ such that the identity doesn't hold. – saz Jul 30 '15 at 15:39
• but if parallelogram law holds then the norm can be induced by inner product and thus the space being considered is also the inner product space. In my case $L^1$ IS the inner product space and in yours - it is not. – Ruth90 Jul 30 '15 at 15:44

By linearity one has: $$f=g:\quad\left(\int|f+g|\mathrm{d}\mu\right)^2+\left(\int|f-g|\mathrm{d}\mu\right)^2=2\left(\int|f|\mathrm{d}\mu\right)^2+2\left(\int|g|\mathrm{d}\mu\right)^2$$ $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|\mathrm{d}\lambda=\int|f|\mathrm{d}\lambda+\int|g|\mathrm{d}\lambda$$

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

Then one obtains: $$\|f+f_\varepsilon\|_1^2+\|f-f_\varepsilon\|_1^2\neq2\|f\|_1^2+2\|f_\varepsilon\|_1^2$$

See the thread: Parallelogram Law