What is the justification for this inequality?

[quasar987]

What is the justification for this inequality?

$$\int_0^1(|u(s)|+|v(s)|)(|u(s)|-|v(s)|)ds\leq \left(\int_0^1(|u(s)|^2+|v(s)|^2)ds\right)^{1/2}\left(\int_0^1(|u(s)-v(s)|^2)ds\right)^{1/2}$$

where u and v are complex-valued square-integrable Riemann integrable functions on [0,1].

Thx.

 

[bham10246]

This reminds me of Holder's inequality, and the right hand side of the inequality is the L^2 norm. Go to Wikipedia and given p and q, your p and q equal 2.

 

[tacman]

The justification for this inequality can be seen through the use of the Cauchy-Schwarz inequality, which states that for two vectors u and v in a vector space, the following inequality holds:

|u∙v| ≤ ||u||∙||v||

where ||u|| and ||v|| are the norms (lengths) of the vectors u and v, respectively.

In this case, we can view the functions u and v as vectors in a vector space, and the integral as the dot product between them. The left side of the inequality can be rewritten as:

∫0^1|u(s)|∙|u(s)-v(s)|ds + ∫0^1|v(s)|∙|u(s)-v(s)|ds

which can be seen as the dot product between the vectors |u(s)| and |u(s)-v(s)|, and between |v(s)| and |u(s)-v(s)|, respectively.

Using the Cauchy-Schwarz inequality, we can then rewrite the left side as:

∫0^1|u(s)|∙|u(s)-v(s)|ds + ∫0^1|v(s)|∙|u(s)-v(s)|ds ≤ (∫0^1|u(s)|^2ds)^1/2∙(∫0^1|u(s)-v(s)|^2ds)^1/2 + (∫0^1|v(s)|^2ds)^1/2∙(∫0^1|u(s)-v(s)|^2ds)^1/2

which is equivalent to the right side of the given inequality.

Therefore, the justification for this inequality lies in the application of the Cauchy-Schwarz inequality to the integral of the product of two functions.