Metric space question and solution

Reference: I had blogged this example earlier, but I myself could not fill in the missing gaps at that time. I am trying again with the help of MathWorld Wolfram and of course, the classic, Introductory Real Analysis by Kolmogorov and Fomin, from which it is picked up for my study.

Consider the set C_{[a,b]} of all continuous functions defined on the closed interval [a,b]. Let the distance function (or metric) be defined by the formula:

\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{1/2} ——– relation I

The resulting metric space will be denoted by C_{[a,b]}^{2}.

The first two properties of a metric space are clearly satisfied by the above function. We need to check for the triangle inequality:

Now I satisfies the triangle inequality because of the following Schwarz’s inequality:

(\int_{a}^{b}x(t)y(t)dt)^{2} \leq \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y(t)dt —— relation II

In order to get to the above relation II, we need to prove the following:

Prove: (\int_{a}^{b} x(t)y(t)dt)^{2} = \int_{a}^{b}x^{2}(t)dt \int_{a}^{b} y^{2}(t)dt - \frac{1}{2}\int_{a}^{b} \int_{a}^{b}[x(s)y(t)-x(t)y(s)]^{2}dsdt.

From the above, we can deduce Schwarz’s inequality (relation II here in this blog article).

(My own attempts failed to crack it so I had to look at the internet for help. Fortunately, MathWorld Wolfram has given a crisp clear proof…but some parts of the proof are still out of my reach…nevertheless, I am reproducing the proof here for the sake of completeness of my notes…for whatever understanding I can derive at this stage from the proof…):

CITE THIS AS:

Weisstein, Eric W. “Schwarz’s Inequality.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/SchwarzsInequality.html

Schwarz’s Inequality:

Let \Psi_{1}(x), \Psi_{2}(x) be any two real integrable functions in [a,b], then Schwarz’s inequality is given by :

|< \Psi_{1}, \Psi_{2}>|^{2} \leq < \Psi_{1}|\Psi_{2}> <\Psi_{2}|Psi_{1}>

Written out explicity,

|\int_{a}^{b} \Psi_{1}(x), \Psi_{2}(x)|^{2} \leq \int_{a}^{b}[\Psi_{1}(x)]^{2}dx \int_{a}^{b}[\Psi_{2}(x)]^{2}dx

with equality if and only if \Psi_{1}(x) = \alpha \Psi_{2}(x) with \alpha a constant. Schwarz’s inequality is sometimes also called the Cauchy-Schwarz inequality or Buniakowsky’s inequality.

To derive the inequality, let \Psi(x) be a complex function and \lambda a complex constant such that

\Psi(x) \equiv f(x) + \lambda g(x) for some f and g

Since

\int \overline{\Psi} \Psi dx \geq 0, where \overline{z} is the complex conjugate.

\int \overline{\Psi}\Psi dx = \int \overline{f}f dx + \lambda \int \overline{f} g dx + \overline\lambda \int \overline{g} f dx + \lambda \overline{\lambda} \int \overline{g} g dx

with equality when \Psi(x) = 0

Writing this, in compact notation:

<\overline{f},f> + \lambda <\overline{f},g> + \overline{\lambda} <\overline{g},f> + \lambda \overline{\lambda} <\overline{g},g> \geq 0….relation A

Now, define \lambda \equiv - \frac{<\overline{g}, f>}{<\overline{g},g>}….relation B

and \overline{\lambda} = - \frac{<g, \overline{f}>}{<\overline{g}, g>}…relation C

Multiply A by <\overline{g},g> and then plug in B and C to obtain:

<\overline{f}, f><\overline{g}, g> - <\overline{f},g><\overline{g},f> - <\overline{g},f><g, \overline{f}> +<\overline{g}, f><g, \overline{f}> \geq 0

which simplifies to

<\overline{g},f><\overline{f},g> \leq <\overline{f},f><\overline{g},g>

So

|<f,g>|^{2} \leq <f,f><g,g>. Bessel’s inequality follows from Schwarz’s inequality. QED.

Regards,

Nalin Pithwa