Operator Zero iff Inner Product Zero

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Theorem

Let $\HH$ be a Hilbert space over $\C$.

Let $A: \HH \to \HH$ be a bounded linear operator.


Suppose that:

$\forall h \in \HH: \innerprod {A h} h_\HH = 0$


Then $A$ is the zero operator.


Proof

For each $x, y \in \HH$ we have:

\(\ds 0\) \(=\) \(\ds \innerprod {\map A {x + y} } {x + y} - \innerprod {\map A {x - y} } {x - y} + i \innerprod {\map A {x + i y} } {x + i y} - i \innerprod {\map A {x - i y} } {x - i y}\) since $\innerprod {A h} h_\HH = 0$ for each $h \in \HH$
\(\ds \) \(=\) \(\ds \innerprod {A x} x + \innerprod {A x} y + \innerprod {A y} x + \innerprod {A y} y - \paren {\innerprod {A x} x - \innerprod {A x} y - \innerprod {A y} x + \innerprod {A y} y}\)
\(\ds \) \(+\) \(\ds i \paren {\innerprod {A x} x + \innerprod {A x} {i y} + \innerprod {i A y} x + \innerprod {i A y} {i y} } - i \paren {\innerprod {A x} x - \innerprod {i A y} x + \innerprod {A x} {i y} - \innerprod {i A y} {i y} }\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \innerprod {A x} y + \innerprod {A y} x - \paren {-\innerprod {A x} y - \innerprod {A y} x} + i \paren {\innerprod {A x} {i y} + \innerprod {i A y} x + \cmod i^2 \innerprod {A y} y} - i \paren {\innerprod {i A y} x + \innerprod {A x} {i y} - \cmod i^2 \innerprod {A y} y}\) Inner Product is Sesquilinear and again using the hypothesis $\innerprod {A h} h_\HH = 0$
\(\ds \) \(=\) \(\ds 2 \innerprod {A x} y + 2 \innerprod {A y} x + i \paren {-i \innerprod {A x} y + i \innerprod {A y} x} - i \paren {i \innerprod {A y} x - i \innerprod {A x} y}\) using the hypothesis $\innerprod {A h} h_\HH = 0$
\(\ds \) \(=\) \(\ds 2 \innerprod {A x} y + 2 \innerprod {A y} x + \innerprod {A x} y - \innerprod {A y} x - \innerprod {A y} x + \innerprod {A x} y\)
\(\ds \) \(=\) \(\ds 4 \innerprod {A x} y\)

So:

$\innerprod {A x} y = 0$

for each $x, y \in \HH$.

Setting $y = A x$, we have:

$\norm {A x}^2 = 0$

by the definition of the inner product norm.

So from Norm Axiom $\text N 1$: Positive Definiteness, we have $A x = 0$ for each $x \in \HH$.

$\blacksquare$


Also see


Sources

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