91Ó°ÊÓ

Positive definiteness is a crucial property that helps to verify whether a given function indeed defines a norm on a vector space. In simple terms, this property ensures that the norm is always non-negative and is zero only for the zero vector.

To illustrate this, consider the context of our exercise: we are given a function norm defined over a space of continuous functions, \( V=\mathscr{C}[0,1] \). For any function \( f(x) \) within this space, the positive definiteness condition requires that \( \int_{0}^{1} |f(x)| \, dx \geq 0 \).

Here's why this works: the integral of \(|f(x)|\), a non-negative function, is naturally non-negative. Furthermore, this integral resulting in zero is possible only when \(|f(x)| \) itself is zero almost everywhere, leading us to conclude that \( f(x) = 0 \) for all \( x \in [0,1] \). Therefore, positive definiteness essentially confirms the presence of a genuine norm by ensuring it only vanishes for functions that are zero everywhere.

Scalar homogeneity, often termed as homogeneity of degree one, is a straightforward yet essential characteristic of norms. It ensures that multiplying a vector by a scalar scales the norm by the absolute value of that scalar. Mathematically, for any scalar \( c \) and vector (or function) \( f \), the property is expressed as \( |cf| = |c| |f| \).

In the problem at hand, we apply this property to functions: \( |cf| = \int_{0}^{1} |c f(x)| \, dx = |c| \int_{0}^{1} |f(x)| \, dx = |c| |f| \). The reasoning is relatively straightforward, as multiplying the function \( f(x) \) by \( c \) results in scaling each value by \( |c| \), thus scaling the integral. 

Itemizing the process:

• For any function \( f(x) \), calculate \(|cf|\).
• The scaling factor \(|c|\) applies uniformly across the function \(f(x)\).
• The final outcome respects the property of scalar homogeneity, demonstrating that the function norm behaves predictably under scalar multiplication.

The triangle inequality is a key feature of norms anchoring them in geometric intuition. It helps illustrate how the `length' of a sum of vectors (or functions) is less than or equal to the sum of their `lengths'.

In our specific context, the triangle inequality for the given function norm implies that the `length' or integral of the sum of two functions \( f \) and \( g \) is no greater than the sum of their `individual lengths'.
The expression goes: \( |f + g| \leq |f| + |g| \). We establish this for functions as follows:

• Find the integral of the absolute value of \( f(x) + g(x) \), expressed as \( \int_{0}^{1} |f(x) + g(x)| \, dx \).
• Utilize the \'real\' triangle inequality property for each point: \( |f(x) + g(x)| \leq |f(x)| + |g(x)| \).
• Integrate to get: \( \int_{0}^{1} |f(x) + g(x)| \, dx \leq \int_{0}^{1} |f(x)| \, dx + \int_{0}^{1} |g(x)| \, dx\).
• The conclusion, \( |f + g| \leq |f| + |g| \), is a direct application of these integrals, showcasing the fidelity of the triangle inequality in function norms.