This upper limit is wrong I need a help

[norbert]

this upper limit is wrong...! I need a help

I have some concerns about this text:


it can easily be shown that F satisfies all the criteria of a vector space. As an example, we demostrate that if


$$\psi_{1}(r)$$ and $$\psi_{2}(r)$$ $$\in$$ F

both can be cosidered to be complex functions
then:

$$\psi(r)$$ $$= \lambda_{1}\psi_{1}(r) + \lambda_{2}\psi_{2}(r)$$ $$\in$$ F (A-1)


where $$\lambda_{1}$$ and $$\lambda_{2}$$ are two arbitrary complex numbers

In order to show that $$\psi(r)^{2}$$ is square integrable

expand $$\psi(r)^{2}$$ :

$$\psi(r)^{2}$$ $$= |\lambda_{1}|^{2}|\psi_{1}(r)|^{2} + |\lambda_{2}|^{2} |\psi_{2}(r)|^{2} + \lambda_{1}^\ast\lambda_2\psi_1^\ast(r)\psi_2(r) + \lambda_1\lambda_2^\ast\psi_1(r)\psi_2^\ast(r)$$ (A-2)

(*) simbol means complex-conjugate

The last two terms of (A-2) have the same modulus, which has as an upper limit:

----- $$|\lambda_{1}||\lambda_{2}| . [ |\psi_{1}(r)|^{2} + |\psi_{2}(r)|^{2}]$$ -----


$$|\psi(r)|^{2}$$ is therefore smaller than a function whose integral converges, since $$\psi_{1}(r)$$ and $$\psi_{2}(r)$$ are square-integrable

The questions are:

Why has the author used the above expression as "upper limit" term?

How did he obtain this expression?

What is the relation of this question with "triangular inequality" referred to complex-variable?

see --- Churchill complex variables and applications ----
can anybody explain me it a little better?

all suggestions will be welcome



thank you all

 

[adriank]

As for why: He used that to show that each term is integrable, by showing that |ψ(r)|² is smaller than a combination of integrable functions, so that |ψ(r)|² is integrable.

As for how: That follows from the fact that 2ab ≤ a² + b². (This can be proven by noting that 0 ≤ (a - b)² = a² + b² -2ab.) In this case, a = |ψ₁(r)| and b = |ψ₂(r)|. I don't see any immediate connection with the triangle inequality.

 

[norbert]

Hi adriank
I am grateful for your help
really there is no connection with the triangle inequality
I can set the last two terms of (A-3) equation to constant a and b respectively
hence using the 2ab < a^2 + b^2 relation I can obtain the upper limit:

$$|\lambda_{1}||\lambda_{2}| . [ |\psi_{1}(r)|^{2} + |\psi_{2}(r)|^{2}]$$

since these terms have the same modulus

thank you