Question

Assume that \(Y_{i}\) are subspaces of a Banach space \(X\) such that \(X=\) \(Y_{1} \oplus Y_{2}\) (algebraic sum). Let \(P_{i}\) be the associated linear projections onto \(Y_{i}\) (so \(\left.P_{1}+P_{2}=I_{X}\right)\). Show that: (i) \(P_{1}(X)=\operatorname{Ker}\left(P_{2}\right)\). (ii) \(P_{1}\) is bounded if and only if \(P_{2}\) is bounded. (iii) Both \(Y_{1}\) and \(Y_{2}\) are closed if and only if both \(P_{i}\) are bounded. Find an example when \(Y_{1}\) is closed but \(Y_{2}\) is not. The complement of a subspace \(Y\) is sometimes defined as a subspace \(Z\) such that \(X=Y \oplus Z\) and the corresponding projections are bounded. (iii) shows that it is an equivalent definition, and the previous exercise shows that closedness of \(Y\) is necessary for the existence of a complement.

Short answer

(i) \( P_1(X) = \operatorname{Ker}(P_2) \). (ii) \( P_1 \) is bounded if and only if \( P_2 \) is bounded. (iii) Both \( P_i \) are bounded if and only if both \( Y_i \) are closed.

Step-by-step solution

Understand the Direct Sum and Projections

Given that the Banach space \( X \) can be written as \( X = Y_1 \oplus Y_2 \), it means any element \( x \in X \) can be uniquely written as \( x = y_1 + y_2 \) where \( y_1 \in Y_1 \) and \( y_2 \in Y_2 \). We have projections \( P_1 \) and \( P_2 \) such that \( P_1(x) = y_1 \) and \( P_2(x) = y_2 \), satisfying \( P_1 + P_2 = I_X \).

Show that \( P_1(X) = \operatorname{Ker}(P_2) \)

By definition, \( P_1(x) = y_1 \) and \( P_2(x) = y_2 \) with \( x = y_1 + y_2 \). To show \( P_1(X) = \operatorname{Ker}(P_2) \), take \( x \in \operatorname{Ker}(P_2) \), then \( P_2(x) = 0 \). This means \( x = P_1(x) \), hence \( x \in P_1(X) \). Conversely, if \( y \in P_1(X) \), then \( y = P_1(x) \) for some \( x \in X \). Also, since \( y \) is in \( P_1 \)'s image, \( P_2(y) = P_2(P_1(x)) = 0 \). Therefore, \( y \in \operatorname{Ker}(P_2) \). Thus, \( P_1(X) = \operatorname{Ker}(P_2) \).

Prove \( P_1 \text{ is bounded if and only if } P_2 \text{ is bounded} \)

Assume \( P_1 \) is bounded. Then since \( P_2 = I_X - P_1 \), \( P_2 \) is the difference of bounded operators (the identity operator \(I_X\) is bounded), so \( P_2 \) is also bounded. Conversely, if \( P_2 \) is bounded, then \( P_1 = I_X - P_2 \) is the difference of bounded operators, so \( P_1 \) is bounded. Thus, \( P_1 \) is bounded if and only if \( P_2 \) is bounded.

Verify That Both \( Y_1 \) and \( Y_2 \) are Closed if and only if Both \( P_i \)'s are Bounded

If both \( P_1 \) and \( P_2 \) are bounded, \( P_1 \) and \( P_2 \) are continuous projections. Continuous projections onto subspaces imply that these subspaces are closed. Therefore, \( Y_1 \) and \( Y_2 \) must be closed when \( P_1 \) and \( P_2 \) are bounded. Conversely, if \( Y_1 \) is closed, \( P_1 \) is bounded; similarly, if \( Y_2 \) is closed, \( P_2 \) is bounded. Hence, both subspaces are closed if and only if both projections are bounded.

Find an Example of \( Y_1 \) Closed but \( Y_2 \) Not Closed

Consider the Banach space \( X = l^p \) for \( p \geq 1 \), and let \( Y_1 \) be sequences eventually zero and \( Y_2 \) be sequences with only finitely many non-zero terms. \( Y_1 \) is a closed subspace (it’s formed by convergence in \( l^p \)), but \( Y_2 \) is not closed since sequences can converge to elements in \( Y_2 \) that are not in \( Y_2 \).

Key concepts

Direct Sum Decomposition

In Banach spaces, the notion of direct sum decomposition allows us to break down a complex space into simpler parts. Given a Banach space \(X\) and its subspaces \(Y_1\) and \(Y_2\), if we can represent any element \(x \in X\) as \(x = y_1 + y_2\), where \(y_1 \in Y_1\) and \(y_2 \in Y_2\), we say that \(X = Y_1 \oplus Y_2\). This decomposition facilitates understanding and solving more complex problems. The unique representation ensures that there's no overlap between the subspaces, making their interaction clear and manageable.
For example, in our exercise, any element in \(X\) can be written as a sum of elements from \(Y_1\) and \(Y_2\). This clarity helps in defining and understanding projections and other related operators.

Linear Projections

Projections are linear operators used to 'project' elements from a larger space onto a subspace. In the context of Banach spaces, if \(X = Y_1 \oplus Y_2\), there exist linear projections \(P_1\) and \(P_2\). These projections satisfy:\[ P_1(x) = y_1 \quad \text{and} \quad P_2(x) = y_2 \]\This means \(P_1(x)\) extracts the part of \(x\) from \(Y_1\) and \(P_2(x)\) does the same for \(Y_2\). The condition \(P_1 + P_2 = I_X\) ensures that the projections collectively reconstruct the original element in \(X\).
For the exercise, the relationship \(P_1(X) = \operatorname{Ker}(P_2)\) indicates that the image of one projection is precisely the kernel of the other—meaning \(P_1\) and \(P_2\) are intimately related.

Bounded Operators

An operator in a Banach space is bounded if there exists a constant \(C\) such that for all elements \(x\) in the space, the operator does not amplify the element's norm beyond \(C\). Mathematically, an operator \(T\) is bounded if: \[ \|T x \| \leq C \| x \| \] for some \(C > 0\) and all \(x \in X\).
In direct sum decomposition, if one of the projections is bounded, the other must be bounded as well. This interplay ensures that both projections manage their respective parts smoothly without causing disruptions in the overall structure of \(X\).
For instance, if \(P_1\) is bounded, then \(P_2\) must also be bounded because their sum is the identity operator \(I_X\), which is inherently bounded.

Closed Subspaces in Banach Spaces

A subspace \(Y\) of a Banach space \(X\) is closed if it contains all its limit points, i.e., the limit of any convergent sequence within \(Y\) still lies in \(Y\). Closed subspaces are important because they guarantee certain properties, such as the ability to define continuous projections.
In our context, if \(P_1\) and \(P_2\) are bounded projections, \(Y_1\) and \(Y_2\) must be closed. This happens because bounded projections are continuous, and the continuity of projections onto subspaces implies these subspaces are closed.
Moreover, the example given in the exercise illustrates a situation where one subspace is closed, and the other is not, providing insight into different behaviors within Banach spaces.

Kernel of Linear Operators

The kernel (or null space) of a linear operator \(T\) in a Banach space is the set of all elements that \(T\) maps to zero. Formally, if \(T: X \to Y\) is a linear operator, the kernel is defined as: \[\operatorname{Ker}(T) = \{ x \in X : T(x) = 0 \}\].
In the exercise, showing that \(P_1(X) = \operatorname{Ker}(P_2)\) means that the image of \(P_1\) is exactly the set of elements that \(P_2\) sends to zero. This relationship helps simplify and understand the structure within \(X\). Kernels highlight elements unaffected by certain projections, enabling deeper insights into the space’s decomposition.
Understanding the kernel's behavior aids in comprehending the operator's overall impact and its nullifying effects within specific subspaces.