Operator with Zero Numerical Range is Zero Operator

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Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.

Let $\norm {\, \cdot \,}$ be the inner product norm on $\struct {\HH, \innerprod \cdot \cdot}$.

Let $\struct {\map D T, T}$ be a densely-defined linear operator on $\HH$ such that:

$\map W T = \set 0$

where $\map W T$ is the numerical range of $T$.


Then $T = 0$.


Corollary

Let $\DD$ be a dense linear subspace of $\HH$.

Let $\struct {\DD, T}$ and $\struct {\DD, S}$ be densely-defined linear operators on $\HH$ such that:

$\innerprod {T x} x = \innerprod {S x} x$ for each $x \in \DD$.


Then $T = S$.


Proof

Let $x, y \in \map D T$.

Since $\map W T = \set 0$, we have:

$\innerprod {\map T {x + y} } {x + y} = \innerprod {T x} x = \innerprod {T y} y = 0$

From Inner Product is Sesquilinear, we have:

$\innerprod {T x} x + \innerprod {T x} y + \innerprod {T y} x + \innerprod {T y} y = 0$

That is:

$\innerprod {T x} y = -\innerprod {T y} x$

Similarly, replacing $y$ with $i y$ and applying Inner Product is Sesquilinear, we have:

$-i \innerprod {T x} y = -\innerprod {\map T {i y} } x$

Since $T$ is linear, we have:

$\innerprod {T x} y = \innerprod {T y} x$

Hence:

$\innerprod {T x} y = -\innerprod {T x} y$

This gives:

$\innerprod {T x} y = 0$ for all $x, y \in \map D T$.

That is:

$T x \in \map D T^\bot$ for all $x \in \map D T$

where $\map D T^\bot$ is the orthocomplement of $\map D T$.

From Linear Subspace Dense iff Zero Orthocomplement, we have $\map D T^\bot = \set 0$.

Hence:

$T x = 0$ for all $x \in \map D T$.

$\blacksquare$


Sources

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