Question

Show that if \(U_{X}=\\{x \in X:\|x\| \leq 1\\}\) and \(T\left(U_{X}\right)\) is bounded, then \(T\) is continuous. [Hint: Take \(x \neq \theta\) and consider \(x /\|x\|\).]

Short answer

Question: Show that if \(U_X = \{x \in X: \|x\| \leq 1\}\) and \(T(U_X)\) is bounded, then the operator \(T\) is continuous. Answer: Given that \(T(U_X)\) is bounded, we have a constant \(M\) such that \(\|T(x)\| \leq M \|x\|\) for all \(x \in U_X\). Now, let \(x \in X\) be a non-zero vector and consider the vector \(y = \frac{x}{\|x\|}\), which is in \(U_X\). Then, we find that \(\|T(y)\| = \|T\left(\frac{x}{\|x\|}\right)\| \leq M\). Using this bound, we can show that \(\|T(x)\| \leq M \|x\|\) for any \(x \in X\), which implies that the operator \(T\) is continuous.

Step-by-step solution

Define the constant of boundedness of T

Let \(M\) be the constant of boundedness of \(T\), meaning \(\|T(x)\| \leq M \|x\|\) for all \(x \in U_X\). Note that this constant exists due to the assumption that \(T(U_X)\) is bounded.

Consider an arbitrary non-zero vector

Now, let \(x \in X\) be a non-zero vector. We want to consider the vector \(y = \frac{x}{\|x\|}\), which is the normalization of \(x\). From the definition of \(U_X\), we know that \(y \in U_X\), as \(\|y\| = 1\).

Calculate the norm of T(y)

Now, calculate the norm of \(T(y)\): \[\|T(y)\| = \|T\left(\frac{x}{\|x\|}\right)\| \leq M \left\|\frac{x}{\|x\|}\right\| = M.\]

Relate the norm of T(y) to the norm of T(x)

From step 3, we have \(\|T(y)\| \leq M\) which is equivalent to \(\|T\left(\frac{x}{\|x\|}\right)\| \leq M\). Using this inequality, we can now express the norm of \(T(x)\) in terms of the norm of \(T(y)\) as follows: \[\|T(x)\| = \left\|T\left(\|x\| \cdot \frac{x}{\|x\|}\right)\right\| = \|x\| \cdot \left\|T\left(\frac{x}{\|x\|}\right)\right\| \leq M \|x\|.\]

Conclude that T is continuous

Now, we have shown that \(\|T(x)\| \leq M \|x\|\) for any \(x \in X\), which means that \(T\) is a continuous linear operator, as claimed.

Key concepts

Bounded Linear Operator

Understanding bounded linear operators is crucial when studying functional analysis, a branch of mathematics dealing with function spaces and linear operators between them. In simple terms, a linear operator, which we may call T, is a rule that assigns to each element of a vector space another vector in a consistent linear fashion -- meaning, T(ax + by) is always equal to aT(x) + bT(y), where x and y are vectors, and a and b are scalars.

A bounded linear operator is one that essentially keeps vectors within limits, preventing them from being stretched to infinity when the operator is applied. Formally, if there is a constant M such that for every vector x, the norm of T(x) doesn't exceed M multiplied by the norm of x, then the operator is considered bounded. This property is directly related to the concept of continuity; a bounded linear operator on a normed vector space is always continuous.

In the exercise provided, we see an illustration of this concept. By showing that the transformed unit ball T(U_X) is bounded, we can deduce that the operator T does not stretch vectors too far out and thus maintains continuity. It's the foundation that ensures the stability of functions and sequences under linear transformations.

Normed Vector Space

A normed vector space is a context in which the concept of 'size' or 'length' applies to vectors. It's a vector space equipped with a function called a 'norm', which measures the magnitude of vectors in the space. The norm is denoted by two vertical bars, like this: \(\|x\|\) for a vector x. This norm must satisfy certain conditions to be valid: it must be non-negative, equal to zero only for the zero vector, and comply with the triangle inequality and scalar multiplication.

When we talk about boundedness or continuity of linear operators, we are dealing with vectors within some normed vector space. In the exercise, U_X represents the unit ball in a normed vector space X, which is the set of all vectors with a norm less than or equal to one. This concept is essential as it allows us to discuss functions that preserve or distort the 'sizes' of vectors through linear operations.

Operator Norm

The operator norm of a linear operator is a measure that tells us how 'strong' or 'intense' an operator is in terms of how much it can stretch vectors. It is defined as the largest amount a linear operator T can stretch a vector, as compared to the vector's original length. In other words, it's the maximum of the ratios \(\|T(x)\|/\|x\|\) over all non-zero vectors x.

Mathematically, the operator norm is denoted and defined as \(\|T\| = \sup\{\|T(x)\| : x \in X, \|x\| = 1\}\), where X is a normed vector space. This definition aligns with the constant M we've seen in the exercise, which caps the norm of T(x) by M*\|x\| for all x. If such an M exists, T is said to have a bounded operator norm. A crucial aspect of operator norms is that if an operator has a finite operator norm, it is bounded, and hence continuous, fitting the larger jigsaw puzzle of linear functional analysis.