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 $S \circ T \ne I$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations