Answered: "Comparison of L2 and L1 norms for functions..."

berljivx8

Answered question

2022-01-14

Comparison of $L^{2} and L^{1}$ norms for functions
In the context of Lebesgue Integrals, I have came across $L^{2}$ as the set of measurable functions $f : [a, b] \to C$ that have Lebesgue integrable squares - that is $x \to {| f (x) |}^{2}$ is Lebesgue integrable.

Answer & Explanation

deginasiba

Beginner2022-01-15Added 31 answers

Yes, the inequality is given by Cauchy-Schwarz:

{| | f | |}_{L^{1}} = \int_{a}^{b} | f (x) | d x \leq {(\int_{a}^{b} {| f (x) |}^{2} d x)}^{\frac{1}{2}} {(\int_{a}^{b} d x)}^{\frac{1}{2}} = \sqrt{b - a} {| | f | |}_{L^{2}}

Jenny Bolton

Beginner2022-01-16Added 32 answers

Hint: Try using the Hölder inequality for integrals on the function

f \cdot 1

.
In general this will not be true. It is of great importance that your measure space (here: [a,b]) has finite measure. Then the

L^{p}

spaces fit nicely into each other, i.e.

L^{q} \subseteq L^{p} whenever p \leq q

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