Question

Let \(T\) be a linear operator (not necessarily bounded) from a normed space \(X\) into a normed space \(Y .\) Show that the following are equivalent: (i) \(T\) is an open map. (ii) There is \(\delta>0\) such that \(\delta B_{Y} \subset T\left(B_{X}\right)\). (iii) There is \(M>0\) such that for every \(y \in Y\) there is \(x \in T^{-1}(y)\) satisfying \(\|x\|_{X} \leq M\|y\|_{Y}\) Hint: (i) \(\Longrightarrow\) (ii): \(T\left(B_{X}^{O}\right)\) is open, so \(y_{0}+\delta B_{Y}^{O} \subset T\left(B_{X}^{O}\right)\) for some \(y_{0} \in Y\), \(\delta>0 . T\left(B_{X}^{O}\right)\) is symmetric and convex, hence every \(b \in \delta B_{Y}^{O}\) belongs to \(T\left(B_{X}^{O}\right)\) since \(b=\frac{1}{2}\left(\left(x_{0}+b\right)-\left(x_{0}-b\right)\right)\) (ii) \(\Longrightarrow(\mathrm{i}): T\left(B_{X}^{O}\right) \supset T\left(\frac{1}{2} B_{X}\right) \supset \frac{\delta}{2} B_{Y} \supset \frac{\delta}{2} B_{X}^{O}\), so \(T\) maps neighborhoods of 0 to neighborhoods of 0 . By linearity, \(T\) is open.

Short answer

All four statements are equivalent: (i) implies (ii), (ii) implies (i), (i) implies (iii), and (iii) implies (i).

Step-by-step solution

Prove (i) implies (ii)

Given that T is an open map, this means that T maps open sets in X to open sets in Y. Consider the open unit ball around zero in X, denoted as \(B_X^O\). Since T is open, T(\(B_X^O\)) is open in Y. Thus, there exists \(y_0 \in Y\) and \(\delta > 0\) such that \(y_0 + \delta B_Y^O \subset T(B_X^O)\). Since \(T(B_X^O)\) is symmetric and convex, any element \(b \in \delta B_Y^O\) can be expressed as \(b = \frac{1}{2}((x_0+b) - (x_0-b))\), implying \(b \in T(B_X^O)\). Thus, we have \(\delta B_Y \subset T(B_X)\).

Prove (ii) implies (i)

Assuming \(\delta B_Y \subset T(B_X)\), consider the sets \(T(\frac{1}{2}B_X)\) and \(\frac{\delta}{2}B_Y\). We have \(T(\frac{1}{2}B_X) \supset \frac{\delta}{2}B_Y\). Setting \(\frac{\delta}{2} = \epsilon\), this implies that \(T\) maps neighborhoods of zero in X to neighborhoods of zero in Y. Since these neighborhoods are mapped linearly, \(T\) must be an open map.

Prove (i) implies (iii)

Given that \(T\) is an open map, the image \(T(B_X^O)\) is an open set in Y containing the origin. Consequently, there exists \(\epsilon > 0\) such that \(\epsilon B_Y^O \subset T(B_X^O)\). For any \(y \in Y\), \(y/\|y\|_Y \in B_Y^O\), thus there exists \(x \in B_X^O\) such that \(T(x) = y/\|y\|_Y \). Consequently, \(T(\|y\|_Y x) = y\), and we have \(\|\|y\|_Y x\|_X \leq M\|y\|_Y\) for some constant M.

Prove (iii) implies (i)

Assume there exists \(M > 0\) such that for every \(y \in Y\), there is \(x \in T^{-1}(y)\) with \(\|x\|_X \leq M\|y\|_Y\). Consider an open set U in X. For every \(y \in T(U)\), \(y/\|y\|_Y \in U\), implying \(T(U)\) is open in Y since it contains neighborhoods of each of its points. Therefore, T is an open map.

Key concepts

Open Mapping Theorem

In functional analysis, the Open Mapping Theorem is a fundamental result that provides important information about the structure of linear continuous operators. It states that if we have a surjective continuous linear operator between two Banach spaces, then it is an open map. This implies that it transforms open sets in the domain space into open sets in the codomain space. This theorem is crucial for understanding the behavior of mappings in infinite-dimensional spaces, influencing the results of many other theorems. Knowing that a linear operator is open can help us derive conclusions about the properties and potential applications of the operator.

Equivalent Conditions

Understanding when a linear operator is an open map involves recognizing certain equivalent conditions. In this specific exercise, three conditions are proved to be equivalent:

• The operator is an open map.
• There is a \( \delta > 0 \) such that \( \delta B_Y \subset T(B_X) \).
• There is an \( M > 0 \) such that for every \( y \in Y \) there exists an \( x \in T^{-1}(y) \) satisfying \( \| x \|_X \leq M \| y \|_Y \).

These conditions allow us to approach the problem from multiple angles. By proving these conditions are equivalent, we gain multiple tools for testing whether a given linear operator is an open map. This versatility is particularly valuable for complex analyses and practical applications.

Normed Spaces

Normed spaces are central to functional analysis. They are vector spaces endowed with a norm, which is a function that assigns a positive length or size to each vector. This norm must satisfy certain properties:

• Non-negativity: \( \| x \| \geq 0 \) for all \( x \in X \), and \( \| x \| = 0 \) if and only if \( x = 0 \).
• Scalar multiplication: \( \| \alpha x \| = | \alpha | \| x \| \) for any scalar \( \alpha \in \mathbb{R} \) or \( \mathbb{C} \).
• Triangle inequality: \( \| x + y \| \leq \| x \| + \| y \| \) for all \( x, y \in X \).

These spaces provide the groundwork for many of theorems and concepts in analysis and applied mathematics. They allow for the generalization of geometry to more abstract contexts, while retaining an intuitive sense of distance and size.

Linear Operators

Linear operators are mappings between vector spaces that preserve vector addition and scalar multiplication. In the context of normed spaces, they are functions \( T : X \rightarrow Y \) such that:

• \