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