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

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

b.1 [imath]\; ||f \pm g|| = ||f|| + ||g||[/imath]

b.2 [imath]\; ||f + g||^{2} + ||f \cdot g||^{2} \, = \, 2(||f|| + ||g||)^{2}[/imath]

b.3 Conclude that the parallelogram law is not satisfied, and therefore, [imath]\mathcal{L}^{1}(\mathbb{R})[/imath] 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 [imath]f = 0[/imath] then [imath]f(x) = 0 \; \; \forall \; \; x[/imath]; It implies that [imath]||f|| = 0[/imath]. Conversely, if [imath]||f|| = 0[/imath] we know for the non-negativity of absolute value that [imath]|f(x)| = 0 \;\; \forall \;\; x[/imath]. Hence [imath]f(x) = 0[/imath] iff [imath]f = 0 \;\;\;\;\; \blacksquare[/imath]

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

I'm so glad for your help!!

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