Continuous Linear Operator over Infinite Dimensional Vector Space is not necessarily Invertible

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Theorem

Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.

Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.

Let $I \in \map {CL} X$ be the identity element.

Let $S, T \in \map {CL} X$.

Suppose the dimension of $X$ is finite:

$d = \dim X = \infty$

Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.


Then $T$ and $S$ are not necessarily invertible.


Proof

Let $X, Y$ be $2$-sequence spaces.

Let $T = L : X \to Y$ be the left shift operator.

Let $S = R : X \to Y$ be the right shift operator.

Let $x := \tuple {a_1, a_2, \ldots} \in X$.

We have that:

\(\ds L \circ R \circ x\) \(=\) \(\ds \map L {\map R {\tuple{a_1, a_2, \ldots} } }\)
\(\ds \) \(=\) \(\ds \map L {\tuple {0, a_1, a_2, \dots} }\) Definition of Right Shift Operator
\(\ds \) \(=\) \(\ds \tuple {a_1, a_2, \ldots}\) Definition of Left Shift Operator
\(\ds \) \(=\) \(\ds I \circ x\) Definition of Identity Element

Hence:

$L \circ R = I$

However:

\(\ds R \circ L \circ x\) \(=\) \(\ds \map R {\map L {\tuple{a_1, a_2, \ldots} } }\)
\(\ds \) \(=\) \(\ds \map L {\tuple {a_2, a_3, \dots} }\) Definition of Left Shift Operator
\(\ds \) \(=\) \(\ds \tuple {0, a_2, a_3, \ldots}\) Definition of Right Shift Operator
\(\ds \) \(\ne\) \(\ds I \circ x\)

Therefore, $T \circ S = I$ but $S \circ T \ne I$.

$\blacksquare$


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