📂Banach Space

Young Operator

Definition

For the normed spaces $X$ and $Y$, the operator $0_{\text{op}} : X \to Y$ below is called the zero operator.

$$ 0_{\text{op}} : x \to 0_{Y}, \quad \forall x \in X $$

$0_{Y}$ is the zero vector of $Y$.

Remark

In ordinary usage one does not distinguish the notation for the zero operator and the zero vector, but here, for clarity, we denote as follows.

• $0_{\text{op}}$: the zero operator
• $0_{Y}$: the zero vector of the vector space $Y$
• $0$: the scalar zero

The zero operator is an element of the set of bounded linear operators $$ B(X, Y) = \left\{ T : X \to Y \mid T \text{ is a bounded linear operator} \right\} $$ and, in particular, is the zero vector in the vector space $B(X, Y)$.

The adjoint operator of a zero operator is also a zero operator. For two zero operators $0_{YX} : X \to Y$ and $0_{XY} : Y \to X$,

$$ \braket{0_{YX}x , y}_{Y} = 0 = \braket{x, 0_{XY}y}_{X} $$

Theorem¹

Let $X$ and $Y$ be inner product spaces, and let $Q : X \to Y$ be a bounded linear operator. Then the following hold.

A necessary and sufficient condition for $(a)$ to be $Q = 0_{\text{op}}$ is that for all $x \in X$ and $y \in Y$, $\braket{Qx, y} = 0$.

For the complex vector space $X$, if $Q : X \to X$ satisfies $\braket{Qx, x} = 0$ for every $x \in X$, then $Q = 0_{\text{op}}$ holds.

Proof

$(a)$

$(\implies)$ implies $\forall y \in Y$ and $\braket{0_{Y}, y} = 0$, so the result is immediate.

$(\impliedby)$ Assume that for all $x \in X$ and $y \in Y$ we have $\braket{Qx, y} = 0$. Then for some $x^{\prime} \in X$ the following holds.

$$ \braket{Qx^{\prime}, y} = 0, \quad \forall y \in Y $$

Then, by the properties of the zero vector, $Qx^{\prime} = 0_{Y}$.

Properties of the zero vector

$$ \forall \mathbf{x}\in X,\ \left\langle \mathbf{x},\mathbf{y} \right\rangle = 0 \implies \mathbf{y}=\mathbf{0} $$

Since $Qx^{\prime} = 0_{Y}$ holds for an arbitrary $x^{\prime}$, we obtain $Q = 0_{\text{op}}$.

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$(b)$

By assumption, for all $v = \alpha x + y \in X$ we have $\braket{Qv, v} = 0$. Expanding this yields:

$$ \begin{align*} 0 &= \braket{Qv, v} \\ &= \braket{Q(\alpha x + y), \alpha x + y} \\ &= |\alpha|^{2}\braket{Qx, x} + \alpha\braket{Qx, y} + \overline{\alpha}\braket{Qy, x} + \braket{Qy, y} \end{align*} $$

By the assumption, the first and the last terms are $0$.

$$ \alpha\braket{Qx, y} + \overline{\alpha}\braket{Qy, x} = 0 $$

This must hold for an arbitrary $\alpha \in \mathbb{C}$, so writing $\alpha = 1$ and $\alpha = i$ gives the following equation.

$$ \braket{Qx, y} + \braket{Qy, x} = 0 $$

$$ \braket{Qx, y} - \braket{Qy, x} = 0 $$

$$ \implies \braket{Qx, y} = 0 \quad \forall x, y \in X $$

By $(a)$, we conclude $Q = 0_{\text{op}}$.

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