Discussion
Let X be the space of all continuous complex valued functions on [0, 1] with inner product
\[ \langle f, g \rangle = \int_{0}^{1} f(t)\overline{g(t)} dt. \]
Let \(T : X \to X\) be defined by \(T(f)(t) = t\, e^{it} f(t)\). Then :
Answer: C
T is normal, but neither self-adjoint nor unitary.
T is multiplication by \(\varphi(t) = te^{it}\). Its adjoint is multiplication by \(\overline{\varphi(t)} = te^{-it}\), since \(\langle \varphi f, g \rangle = \int \varphi f \bar{g}\), which equals \(\langle f, \bar{\varphi} g \rangle\). Multiplication operators commute, so \(TT^{*} = T^{*}T\) (both are multiplication by \(t^{2}\)) and T is normal.
T is not self-adjoint, because \(\varphi\) is not real valued (\(\varphi \neq \bar{\varphi}\)). It is not unitary, because \(TT^{*}\) is multiplication by \(|\varphi|^{2} = t^{2}\), not the identity. So A, B and D are wrong.
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