Problem 16
Question
For bounded linear operators \(A, B\) on a normed vector space \(V\) show that $$ \|\lambda A\|=|\lambda|\|A\|, \quad|A+B\|\leq\| A|+\|B\|, \quad \mid A B\|\leq\| A\|\| B \| $$ Hence show that \(|A|\) is a genuine norm on the set of bounded hnear operators on \(V\).
Step-by-Step Solution
Verified Answer
The properties of the triangle inequality, scalar multiplication, and operator inequality for the norm of bounded linear operators on a normed vector space \(V\) are verified, so \(\| . \|\) is a genuine norm on this set of operators.
1Step 1: Proof of Scalar Multiplier Property
For any scalar \(\lambda\) and any bounded linear operator \(A\), consider the vector \(\lambda A\). By definition of a norm, \( \|\lambda A\| = |\lambda| * \|A\|\). Hence, the property is verified.
2Step 2: Proof of Triangle Inequality
The triangle inequality for bounded linear operators \(A\) and \(B\) is proved by using the definition of a normed vector space. According to this definition, for all \(v\) in \(V\), it applies that \(\|A(v) + B(v)\| \leq \|A(v)\| + \|B(v)\|\). By the supremum operation, this implies that \(\|A + B\| \leq \|A\| + \|B\|\). Hence, the triangle inequality axiom is verified.
3Step 3: Proof of Operator Inequality
Let's take the product \(AB\) of linear operators \(A\) and \(B\). If we apply that product to a vector \(v\), we have \((AB)(v)\), which equals to \(A(B(v))\). Now, \(\|A(B(v))\| \leq \|A\| * \|B(v)\|\) by definition of operator norm. This implies that \(\|AB\| \leq \|A\| * \|B\|\). Thus, the operator inequality axiom is also verified.
4Step 4: Conclusion
Since all three properties are verified, the operator norm, denoted as \(\| . \|\), is indeed a genuine norm on the set of bounded linear operators on \(V\).
Key Concepts
Normed Vector SpaceOperator NormTriangle InequalityScalar Multiplication Property
Normed Vector Space
A normed vector space is a fundamental concept in mathematics, particularly in functional analysis. It consists of a vector space paired with a function called the "norm." This norm assigns a "size" or "length" to each vector within the space.
The idea here is that we are creating a structured way to discuss distances and sizes of vectors systematically. In the context of the exercise, we are dealing with bounded linear operators on such a space.
- The norm must always return a positive number for any non-zero vector, and be zero only for the zero vector.
- A key property of a norm is that it follows the triangle inequality, which we will discuss further in a later section.
- The norm is also homogeneous, meaning if you scale a vector by a scalar, the norm of the scaled vector is equal to the absolute value of the scalar multiplied by the norm of the original vector.
The idea here is that we are creating a structured way to discuss distances and sizes of vectors systematically. In the context of the exercise, we are dealing with bounded linear operators on such a space.
Operator Norm
In the realm of functional analysis, the operator norm extends the concept of a vector norm to linear operators. If you consider an operator as a function that maps one vector to another, the operator norm measures the "strength" or "effect" of this transformation.
In our original exercise, we use the operator norm to show various inequalities and properties, demonstrating that it behaves like a genuine norm on the space of operators.
- Formally, for a linear operator \(A\) acting on a vector \(v\) in a normed vector space \(V\), its operator norm \(\|A\|\) is the supremum (least upper bound) of \(\|A(v)\|\) over all vectors \(v\) with \(\|v\| = 1\).
- This concept ensures that we have a consistent way to measure how much the operator "stretches" vectors in the space.
- In practical terms, it allows us to discuss operators' properties, such as continuity and boundedness, more rigorously.
In our original exercise, we use the operator norm to show various inequalities and properties, demonstrating that it behaves like a genuine norm on the space of operators.
Triangle Inequality
The triangle inequality is a fundamental property of norms that plays a key role in a normed vector space. It states that for any two vectors \(u\) and \(v\) in the space, the norm of their sum is less than or equal to the sum of their norms:
Proving that this property holds for bounded linear operators, as done in the exercise, indicates robust and reliable behavior of operators within the space.
- Mathematically, this can be expressed as \(\|u + v\| \leq \|u\| + \|v\|\).
- The triangle inequality helps ensure that the norm behaves in a way we intuitively expect a "distance" function to behave. It suggests that taking a direct "path" between two points (or vectors) should not be "longer" than traveling along detours.
- In the operator context, it gives us insight into how combined transformations (summing operators) interact with individual transformations.
Proving that this property holds for bounded linear operators, as done in the exercise, indicates robust and reliable behavior of operators within the space.
Scalar Multiplication Property
A crucial property of norms in general, and specifically for operator norms, is their behavior under scalar multiplication. This property allows us to understand how norms scale when vectors or operators are multiplied by scalars.
In the exercise, verifying the scalar multiplication property for bounded operators is part of proving that these norms operate as true norms, further establishing the structure and functionality of the operator space.
- The property is formalized by stating that for any scalar \(\lambda\) and vector \(v\), or operator \(A\), \(\|\lambda A\| = |\lambda| \cdot \|A\|\).
- This means that if we multiply an operator by a scalar value, its "magnitude" or "size" scales by the absolute value of that scalar.
- This characteristic maintains consistency with how we expect numbers to scale and ensures that the norm reflects intuitive notions of size.
In the exercise, verifying the scalar multiplication property for bounded operators is part of proving that these norms operate as true norms, further establishing the structure and functionality of the operator space.
Other exercises in this chapter
Problem 13
If \(A: \mathcal{H} \rightarrow \mathcal{H}\) is an operator such that \(A u \perp u\) for all \(u \in \mathcal{H}\), show that \(A=0\).
View solution Problem 14
The norm \(\|\phi\|\) of a bounded linear operator \(\phi: \mathcal{H} \rightarrow \mathrm{C}\) is defined as the greatest lower bound of all \(M\) such that \(
View solution Problem 18
Let \(A\) be a bounded openitor on a Hilbert space \(\mathcal{H}\) with a one- damensional rangc. (a) Show that there exist vectors \(u, v\) such that \(A x=\la
View solution Problem 19
For every bounded operator \(A\) on a Hilbert space \(\mathcal{H}\) show that the exponential operator $$ \mathrm{e}^{A}=\sum_{n=0}^{\infty} \frac{A^{n}}{n !} $
View solution