Can somebody help me with this proof of inner product in L1(R)?

[User13587]

I want to prove that a function space can be normed without a inner product, for that, I need to prove this in the L1; I'd tried to solve the axioms for a normed space but I don't get any good result, I don't know how to get the proof of this.


Let be $\mathcal{L}^{1}(\mathbb{R})$ the set of all functions such that: $$\||f|| = \int_{-\infty}^{\infty} \, |f(x)| dx$$a) Show that $||\cdot||$ defines a norm for $\mathcal{L}^{1}(\mathbb{R})$

b) Let be $f$ and $g$ two non-zero functions such that at no point $x \in \mathbb{R}$ both are different from zero. Verify that:

b.1 $\; ||f \pm g|| = ||f|| + ||g||$

b.2 $\; ||f + g||^{2} + ||f \cdot g||^{2} \, = \, 2(||f|| + ||g||)^{2}$

b.3 Conclude that the parallelogram law is not satisfied, and therefore, $\mathcal{L}^{1}(\mathbb{R})$ is NOT a space with inner product

 

[blamocur]

I've counted 7 questions there: 4 properties of a normed space and 3 questions under b) -- have you succeeded with any of them? E.g., which of the normed space axioms do you have problem with?

 

[User13587]

For example, this is what I wrote for the non-negativity propertie, but I don't know if it is correct and I don't know how to verify it.

Non-negativity

If $f = 0$ then $f(x) = 0 \; \; \forall \; \; x$; It implies that $||f|| = 0$. Conversely, if $||f|| = 0$ we know for the non-negativity of absolute value that $|f(x)| = 0 \;\; \forall \;\; x$. Hence $f(x) = 0$ iff $f = 0 \;\;\;\;\; \blacksquare$

I'm so glad for your help!!

 

[blamocur]

For example, this is what I wrote for the non-negativity propertie, but I don't know if it is correct and I don't know how to verify it.

Non-negativity

If $f = 0$ then $f(x) = 0 \; \; \forall \; \; x$; It implies that $||f|| = 0$. Conversely, if $||f|| = 0$ we know for the non-negativity of absolute value that $|f(x)| = 0 \;\; \forall \;\; x$. Hence $f(x) = 0$ iff $f = 0 \;\;\;\;\; \blacksquare$

I'm so glad for your help!!

I am rusty on these topics, but, technically, one can have a non-zero function for which $\int |f(x)| \neq 0$, but such function cannot be continuous.