Question

Find a discontinuous linear map \(T\) from some Banach space \(X\) into \(X\) such that \(\operatorname{Ker}(T)\) is closed. Hint: Let \(X=c_{0}\) and \(T(x)=\left(f(x), x_{1}, x_{2}, \ldots\right)\) for \(x=\left(x_{i}\right)\), where \(f\) is a discontinuous linear functional on \(X\).

Short answer

A discontinuous linear map is \(T(x) = (f(x), x_{1}, x_{2}, \ldots)\) on \(X = c_{0}\) with \(f\) discontinuous and \(\operatorname{Ker}(T)\) closed.

Step-by-step solution

Understand the given Banach space

First, identify the Banach space given in the hint. The space is defined as \(X = c_{0}\), which is the space of all sequences that converge to 0, with the supremum norm \(orm{\bullet}\).

Define the linear map \(T\)

Consider the linear map \(T: X \to X\) defined by \(T(x) = (f(x), x_{1}, x_{2}, \ldots)\) for \(x = (x_{i})\). Here, \(f\) is a discontinuous linear functional on \(X\).

Identify the kernel of \(T\)

The kernel of the map \(T\) is the set of all sequences \((x_i)\) in \(X\) such that \(f(x) = 0\). This can be written as \(\operatorname{Ker}(T) = \{(x_i) \in c_0 : f(x) = 0\}\).

Verify the kernel is closed

Since \(f\) is linear and continuous, the set of all sequences for which \(f(x)=0\) forms a closed subspace of \(X=c_{0}\). This is because the preimage of a closed set under a continuous linear map is closed.

Verify the map is discontinuous

The functional \(f\) was given as discontinuous, making \(T\) a discontinuous map. This is because the first component of \(T(x)\), which is \(f(x)\), is discontinuous.

Key concepts

discontinuous linear map

In the realm of functional analysis, a linear map is a function that preserves vector addition and scalar multiplication. This means if you have two vectors, say \(u\) and \(v\), and a scalar \(a\), then a linear map \(T\) would satisfy the properties: \(T(u + v) = T(u) + T(v)\) and \(T(a \, u) = a \, T(u)\). Now, a discontinuous linear map doesn’t behave *nicely* with respect to convergence. That is, sequences that should converge nicely after applying the map might not. Consider a Banach space \(X = c_0\). This space consists of all sequences of scalars that converge to 0, endowed with the supremum norm \(\|\cdot\|_{\infty}\).

To see a discontinuous linear map in action, imagine our linear map \(T: X \to X\) defined as \(T(x) = (f(x), x_1, x_2, \ldots)\). Here, \(f\) is a notorious player—a discontinuous linear functional on \(X\). Since \(f\) is discontinuous, it doesn’t play by the rules of nice, continuous maps. Even if a sequence in \(c_0\) smoothly tends to 0, \(f(x)\) can misbehave, meaning it won't tend to zero no matter how you squeeze \(x\) towards the limit.

Understanding discontinuous maps helps highlight how even small tweaks in linear structure can lead to unexpected behaviors, especially in infinite-dimensional settings like Banach spaces.

kernel of linear map

Next, let’s talk about the kernel (or null space) of a linear map. In simple terms, the kernel of a map \(T\), denoted as \(\operatorname{Ker}(T)\), is the set of all elements that \(T\) sends to zero. Mathematically, these are all the \(x\) in \(X\) such that \(T(x) = 0\). Think of the kernel as the starting point of all vectors which collapse into nothingness under the transformation \(T\).

In our specific example, the linear map \(T\) is given by \(T(x) = (f(x), x_1, x_2, \ldots)\). Therefore, the kernel of \(T\) would be all sequences \((x_i)\) in \(c_0\) for which the aforementioned map equals zero. This is expressed as:
\[ \operatorname{Ker}(T) = \{ (x_i) \in c_0 : f(x) = 0 \} \]

Essentially, this kernel is the set of sequences in \(c_0\) that are annihilated by the discontinuous functional \(f\). Exploring the kernel affords insight into the structure and behavior of \(T\), essentially telling us about the 'blind spots' of the map.

closed subspace

Understanding closed subspaces is key in analyzing linear maps. A subspace of a vector space \(X\) is simply any subset that itself is a vector space. When we say a subspace is 'closed,' we're talking about a property related to limits. Specifically, a subspace \(Y\) of \(X\) is closed if it contains all its limit points. To put it another way, if any sequence in \(Y\) converges to a point in \(X\), that point must also lie within \(Y\).

Why does this matter? Well, in our case, we're examining the kernel of our discontinuous linear map \(T\) and verifying that it is indeed a closed subspace. The approaching point comes from understanding that \(\operatorname{Ker}(T) = \{(x_i) \in c_0 : f(x) = 0\}\) forms a closed subspace of \(c_0\).

This can be further grasped by noting that the functional \(f\), which is linear (even if not continuous), induces this closure. Since being closed under linear operations guarantees that \(\operatorname{Ker}(T)\) retains its properties no matter how tiny the elements become within it. Therefore, in the context of Banach spaces, knowing a kernel is closed helps in applying many useful theorems, like the Closed Graph Theorem, which are pillars in functional analysis.