Counterexample of $\text{Null}(T)^{\bot} = \text{Im}(T^{*})$

Question

I know that $\text{Null}(T)^{\bot} = \text{Im}(T^{*})$, where $T^{*}$ means the adjoint operator of a linear operator $T$, holds when the domain of $T$ is finite-dimensional. However, the proof uses the fact that the adjoint of the adjoint of an operator is itself, which holds only on finite-dimensional vector spaces. Is there a counterexample when I extend the domain of $T$?

Answer 1

Let $X=L^2[0,1]$ and define $T:X \to X$ by $(Tf)(x) = x f(x)$. Then $T$ is bounded and $T^* =T$.

Let $c\neq 0$ be a constant, and let $\gamma(x) = c$ for all $x$. Then we see that $\|Tf-\gamma\|>0$ for all $f \in X$ (since $x \mapsto { c \over x}$ is not in $X$). Hence $\gamma$ is not in the range of $T$.

If we choose $g \in X$, we can define $f_n(x) = \begin{cases} {g(x) \over x}, & x \ge {1 \over n} \\ 0,& \text{otherwise}\end{cases}$. Then $f_n \in X$ for all $n$ and $\|Tf_n-g\| \to 0$, hence the range of $T$ is dense in $X$.

In particular, ${\cal R}T$ is not closed, but the orthogonal complement is always closed, hence ${\cal R}T \neq (\ker T)^\bot$.

Indeed, if $Tf=0$, then $f(x) =0$ ae., hence $f=0$, and so $\ker T = \{0\}$, so $(\ker T)^\bot = X$.