Question

Verify that $$ \left(\int_{a}^{b} x(t) y(t) d t\right)^{2}=\int_{a}^{b} x^{2}(t) d t \int_{a}^{b} y^{2}(t) \mathrm{dt}-\frac{1}{2} \int_{a}^{b} \int_{a}^{b}[x(s) y(t)-y(s) x(t)]^{2} d s \mathrm{dt} $$ Deduce Schwarz's inequality (11) from this identity.

Short answer

The identity ensures that the difference \( \geq 0 \), validating Schwarz's inequality by construction.

Step-by-step solution

Interpret the Identity

We are given the identity \( \left(\int_{a}^{b} x(t) y(t) dt\right)^{2} = \int_{a}^{b} x^{2}(t) dt \int_{a}^{b} y^{2}(t) dt - \frac{1}{2} \int_{a}^{b} \int_{a}^{b}[x(s) y(t) - y(s) x(t)]^{2} ds dt \). The goal is to verify this equation and then use it to deduce Schwarz's inequality.

Consider Scalar Product Definition

Recognize that the integral \( \int_{a}^{b} x(t) y(t) dt \) acts analogous to a scalar product of vectors in the context of function spaces. In this analogy, the square of the integral can be seen as the square of the length of a projection.

Expand the Square Representation

Expand the square on the right-hand side of the identity: \( [x(s) y(t) - y(s) x(t)]^{2} \). This results in: \( x(s)^{2}y(t)^{2} - 2x(s)y(t)y(s)x(t) + y(s)^{2}x(t)^{2} \).

Integrate over the Domain

Integrate the expanded expression over \( s \) and \( t \), and substitute back to check if both sides of the given identity match. Note any relationships and simplifications that arise, particularly around the middle term, which supports the correlation context with the integral of the product \( x(t)y(t) \).

Consider Schwarz's Inequality

Schwarz's inequality is expressed as: \( \left( \int_{a}^{b} x(t) y(t) dt \right)^{2} \leq \left( \int_{a}^{b} x^{2}(t) dt \right) \left( \int_{a}^{b} y^{2}(t) dt \right) \). Identify that subtracting a non-negative quantity from the product integral naturally supports this inequality.

Deduce the Inequality

Since \( \frac{1}{2} \int_{a}^{b} \int_{a}^{b}[x(s) y(t) - y(s) x(t)]^{2} ds dt \geq 0 \), subtracting it from the product of integrals \( \int_{a}^{b} x^{2}(t) dt \int_{a}^{b} y^{2}(t) dt \) guarantees the original integral squared is less than or equal to this product, thus satisfying Schwarz's inequality.

Key concepts

Scalar Product in Function Spaces

In mathematics, we often look for ways to compare and relate functions, much like we do vectors in geometry. A useful tool for this comparison is the **scalar product** (or inner product) in function spaces. Instead of multiplying two numbers or vectors, we add an integral into the mix. If you picture regular vectors with components that we multiply and sum for the dot product, functions like \(x(t)\) and \(y(t)\) use a similar process through integrals.

• To compute a scalar product for functions, we use the integral form: \( \int_{a}^{b} x(t) y(t) \, dt \).
• The integral \( \int_{a}^{b} x(t) y(t) dt \) works like the dot product, measuring how much one function contributes to another over an interval \([a, b]\).
• This concept serves as a foundation in verifying equations involving function spaces such as the one given in the exercise.

Think of this process as a projection: the square of this integral can be compared to the length squared of the projection vector, forming a building block to explore complex inequalities that arise in calculus.

Integral Identities

Integral identities are equations involving integrals that are true for all functions meeting certain criteria. These identities help solve problems and deduce more intricate mathematical principles.In our exercise, the integral identity connects the square of an integral to a combination of other integrals.

• On the left side, we have \( \left( \int_{a}^{b} x(t) y(t) dt \right)^{2} \), illustrating a squared scalar product.
• The right side contains three distinct parts: the product of two squared integrals \( \int_{a}^{b} x^{2}(t) dt \) and \( \int_{a}^{b} y^{2}(t) dt \), minus a double integral \( \frac{1}{2} \int_{a}^{b} \int_{a}^{b} [x(s) y(t) - y(s) x(t)]^{2} ds dt \).

Understanding the relationship in this identity provides insights into functional behaviors and properties, acting as a bridge or story that explains the strength and nuances of interactions between different functions over specified ranges.

Expansion of Squares

The **expansion of squares** lies at the core of many algebraic manipulations, especially in integral identities and inequalities.Consider the expression \( [x(s) y(t) - y(s) x(t)]^{2} \):

• This illustrates a typical expansion: \( x(s)^{2}y(t)^{2} - 2x(s)y(t)y(s)x(t) + y(s)^{2}x(t)^{2} \).
• The expression is expanded into a form where each term can be separately integrated.
• This process helps decompose complex expressions into simpler constituents, facilitating the evaluation of integral identities.

By expanding products into sums, mathematicians and students can more easily compute and verify integral identities. In this way, subtracting non-negative terms, shown in the provided identity, naturally aids in deducing inequalities like Schwarz's. Recognizing these expanded forms and their integrals makes it easier to appreciate the structural balance in such mathematical expressions.