Question

Show that $$ \|T\|=\sup _{\|x\| \mid \leq 1}\|T x\| \text { and }\|T\|=\sup _{\|x\|=1}\|T x\| $$ both define \(\|T\|\).

Short answer

#Short Answer# Both expressions— \(\sup_{\|x\| \leq 1}\|Tx\|\) and \(\sup_{\|x\|=1}\|Tx\|\)—are equal and define the norm of the linear operator T. We demonstrated this by establishing an inequality between these expressions, showing that the first expression is greater than or equal to the second, and vice versa. Since these expressions are equal and they satisfy the conditions of a norm, we can conclude that both define the same norm \(\|T\|\) for the linear operator T.

Step-by-step solution

Understanding the two expressions

We are given two expressions: 1. \(\|T\|=\sup_{\|x\| \leq 1}\|Tx\|\) : The supremum is taken over all possible x with norm less than or equal to 1. 2. \(\|T\|=\sup_{\|x\|=1}\|Tx\|\) : The supremum is taken over all possible x on the unit sphere (i.e., with norm equal to 1). Our goal is to show that both expressions define the norm of the linear operator T.

Establishing an inequality between the two expressions

First, let's show that \(\sup_{\|x\|\leq 1}\|Tx\| \geq \sup_{\|x\|=1}\|Tx\|\). Consider any vector \(x\) with \(\|x\| = 1\). Since \(\|x\|\leq 1\), the value \(\|Tx\|\) is also included in the set \(\{\|Tx\|: \|x\|\leq 1\}\). Thus, the supremum of this set will always be greater than or equal to the supremum of the set \(\{\|Tx\|: \|x\|= 1\}\). Now, let's show that \(\sup_{\|x\|\leq 1}\|Tx\| \leq \sup_{\|x\|=1}\|Tx\|\). Consider any vector \(y\) with \(\|y\| \leq 1\). If \(\|y\| = 1\), then \(\|Ty\|\) is included in the set \(\{\|Tx\|: \|x\|= 1\}\), and if \(\|y\| \neq 1\), we can always scale \(y\) by a factor of \(\frac{1}{\|y\|}\) making a new vector \(x = \frac{1}{\|y\|}y\) such that \(\|x\| = 1\). Then, $$ \|Ty\| = \left\| T\left(\|y\| x\right) \right\| = \|y\|\|Tx\| $$ As \(0 \leq \|y\| \leq 1\), we have \(0 \leq \|y\|\|Tx\| \leq \|Tx\|\), which implies that \(\|Ty\|\) is less than or equal to the supremum of the set \(\{\|Tx\|: \|x\|= 1\}\). Thus, we conclude that $$ \sup_{\|x\|\leq 1}\|Tx\| \leq \sup_{\|x\|=1}\|Tx\|. $$

Concluding the equality and showing both expressions define the norm of T

Since we have demonstrated that: $$ \sup_{\|x\|\leq 1}\|Tx\| \geq \sup_{\|x\|=1}\|Tx\| $$ and $$ \sup_{\|x\|\leq 1}\|Tx\| \leq \sup_{\|x\|=1}\|Tx\|, $$ it follows that $$ \sup_{\|x\|\leq 1}\|Tx\| = \sup_{\|x\|=1}\|Tx\|. $$ Therefore, both expressions are equal, and since they both satisfy the conditions of a norm (non-negativity, linearity, and the triangle inequality), we can conclude that both expressions define the same norm \(\|T\|\) for the linear operator T.

Key concepts

Linear Operators

Linear operators are fundamental objects in mathematics, especially in linear algebra and functional analysis. They are mappings between two vector spaces that preserve the operations of vector addition and scalar multiplication. This means if you have a linear operator \( T \), and vectors \( x \) and \( y \), as well as a scalar \( c \), the following properties are satisfied:
\( T(x + y) = T(x) + T(y) \)
\( T(cx) = cT(x) \).

Linear operators can be thought of as matrices, providing a way to transform vectors within or between spaces in a consistent manner. They play a crucial role in understanding various phenomena in physics, engineering, and computer science. Depending on the context, linear operators can potentially simplify complex problems by retaining the structure of equations, thus allowing for efficient solutions using well-understood mathematical techniques.

• Preserve vector operations
• Similar to matrices
• Essential in mathematical problem solving

Understanding linear operators is key to grasping advanced concepts like transformations, eigenvalues, and eigenvectors. They provide a structured way to map inputs to outputs, making them invaluable in both theoretical and applied mathematics.

Normed Spaces

Normed spaces are a type of vector space equipped with a function called a "norm". This function assigns a non-negative length or size to each vector in the space, which intuitively measures the "distance" of the vector from the origin. A norm \( \| \cdot \| \) must satisfy three important properties:

• Non-negativity: \( \|x\| \geq 0 \) for all vectors \( x \), and \( \|x\| = 0 \) if and only if \( x = 0 \).
• Scalability: \( \|cx\| = |c|\|x\| \) for any scalar \( c \) and vector \( x \).
• Triangle inequality: \( \|x + y\| \leq \|x\| + \|y\| \) for any vectors \( x \) and \( y \).

These properties ensure that the norm behaves much like the Euclidean norm (or "length") that we are familiar with from elementary geometry.

Normed spaces provide the framework required for discussing convergence, continuity, and boundedness, all of which are crucial concepts in analysis. They help in generalizing notions of "length" and "distance" beyond the finite-dimensional spaces, allowing mathematicians to work with infinite-dimensional spaces, which appear frequently in functional analysis. Applications of normed spaces abound in mathematics and science, particularly in solving differential equations, optimization problems, and signal processing.

Supremum

The concept of a supremum is essential in understanding the bounds of sets. In mathematical analysis, the supremum of a set is the smallest real number that is greater than or equal to every number in the set. This idea extends the concept of a maximum, which is the greatest element in a set, although it applies more generally because not all sets have maximum elements.

Using the notation \( \sup \), the supremum is vital in defining norms for operators. When you talk about the operator norm \( \|T\| \), it involves the supremum of certain sets of vector norms. Specifically, the operator norm of \( T \) is given by the largest value of \( \|Tx\| \) over all vectors \( x \) within a certain norm-bound region, for example \( \|x\| \leq 1 \).

The supremum is a "least upper bound", which means it is not necessarily a value contained within the set but is the minimal value that can serve as an upper limit. This makes it a key tool in calculus and analysis when discussing continuity, integrals, and sequences.

• Minimally encases a set
• Handles non-maximum set cases
• Central in defining operator norms

Understanding the supremum helps with comprehending the behavior of functions and operations within certain constraints, as well as ensuring solutions and analyses are sound and comprehensive.