If $f\in C[0,1]$ let $T_f(g)=fg$ for $g\in L_2[0,1]$. Show that $T_f$ is continuous, linear and $(T_f)^\ast=T_{\overline{f}}$.
I showed that $T_f$ is linear and continuous. My question is in the proof of $(T_f)^\ast=T_{\overline{f}}$.
My work:
From the unicity of Adjoint of $T_f$ I need show that for all $g,h\in L_2[0,1]$ and $f\in C[0,1]$ $$\langle T_f(g), h\rangle=\langle g, \overline{f}h\rangle$$ But $$\langle g, T_{\overline{f}}(h)\rangle=\langle g, \overline{f}h\rangle=f\langle g, h\rangle=\langle fg, h\rangle=\langle T_f(g), h\rangle$$
My proof is correct?
My Doubt is:
I can treat $f$ as a scalar and take it out of the inner product like I did before?
You cannot treat $f$ as a scalar.
$\langle T_f g, h \rangle= \int fg\overline {h} =\int g \overline {T_f^{*} h}$ for all $g \in L^{2}$. This implies that $f\overline h=\overline {T_f^{*} h}$ almost everywhere. This is same thing as $T_f^{*} h=h\overline f=T_{\overline f} h$. So $T_f^{*} =T_{\overline f} $.
I have used two facts:
If $F_1$ and $F_2 \in L^{2}$ then $F_1F_2 \in L^{1}$.
If $G_1, G_2 \in L^{1}$ and $\int g G_1=\int gG_2$ for all $g \in C[0,1]$ then $G_1=G_2$ almost everywhere.