Question

Prove that a Banach space \(X\) is injective if and only if for every superspace \(Y\) of \(X\), Banach space \(Z\), and \(T \in \mathcal{B}(X, Z)\) there exists an extension of \(T\) to \(Y\); that is, some \(\widetilde{T} \in \mathcal{B}(Y, X)\) such that \(\left.\widetilde{T}\right|_{X}=T\).

Short answer

A Banach space X is injective if and only if any operator from X to any Z can be extended to any superspace Y of X.

Step-by-step solution

Define Injectivity

A Banach space X is injective if for every Banach space Y containing X as a subspace, and for every Banach space Z, any bounded linear operator T from X to Z can be extended to a bounded linear operator \( \widetilde{T} \) from Y to Z such that \( \left.\widetilde{T}\right|_{X}=T \).

Show Necessity

Assume X is injective. Given a superspace Y of X and a Banach space Z, for any bounded linear operator T in \( \mathcal{B}(X, Z) \), there exists an extension \( \widetilde{T} \in \mathcal{B}(Y, Z) \) such that \( \left.\widetilde{T}\right|_{X}=T \). This follows directly from the definition of injectivity.

Show Sufficiency

Assume for any superspace Y of X, Banach space Z, and bounded linear operator T in \( \mathcal{B}(X, Z) \), there exists an extension \( \widetilde{T} \in \mathcal{B}(Y, Z) \) such that \( \left.\widetilde{T}\right|_{X}=T \). Given any Banach space Y containing X, and any operator T in \( \mathcal{B}(X, Z) \), we can always find such an extension. Hence, X is injective by definition.

Conclusion

The conditions given in the problem statement are both necessary and sufficient for X to be injective: any operator T from X to Z can always be extended to an operator from Y to Z.

Key concepts

Banach space

A Banach space is a complete normed vector space. This means that you can take any sequence of elements from the space that converges and be sure it will converge to an element within the space. For example, consider the space of continuous functions on [0,1] with the maximum norm. If you have a sequence of continuous functions that converges uniformly, the limit is also a continuous function in the space. The completeness property is crucial when dealing with analysis in infinite dimensions.

• **Vector space:** A collection of vectors where you can add any two vectors and multiply by scalars, maintaining closure under these operations.
• **Norm:** A function that assigns a non-negative length or size to all vectors in the space except the zero vector.
• **Completeness:** All Cauchy sequences (sequences where elements get arbitrarily close to each other) in the space converge to a limit within that space.

Bounded linear operator

A bounded linear operator is a linear transformation between two Banach spaces that maintains boundedness. Essentially, it means that the function we are discussing does not stretch vectors arbitrarily large. More formally, a linear operator T from a Banach space X to another Banach space Z is bounded if there exists a constant C such that for all vectors x in X, the norm of T(x) is less than or equal to C times the norm of x, represented as \(\|T(x)\| \leq C \|x\|\).

Some important points:

• **Linearity:** T(a + b) = T(a) + T(b) and T(\(\alpha\)a) = \(\alpha\)T(a) for any vectors a, b, and scalar \(\alpha\).
• **Boundedness:** There is a maximum 'stretch' factor C that applies uniformly to the transformation of vectors in X.
• **Continuous:** A bounded linear operator is always continuous; small changes in the input lead to small changes in the output.

Injective space

An injective Banach space has the special property that every bounded linear operator defined on it can be extended to a larger space without loss of its defining properties. This is a vital concept because it allows for significant flexibility in extending and analyzing functions.

Here's what injectivity involves:

• **Extension:** Every bounded linear operator T from an injective Banach space X to any other Banach space Z can be extended to a superspace Y containing X.
• **Restriction:** The extended operator retains functionality on X, meaning \(\left.\widetilde{T}\right|_{X}=T\).
• **Superspace:** This is a larger space Y that contains X as a subspace.

The key requirement is that you can always find such an extension regardless of what the larger space Y might be. This makes injective spaces highly versatile and useful in functional analysis.