If $Tf = Df$ is the standard differential operator on $\mathbf{R}$, then $T$ acts as a bounded operator from $H^1(\mathbf{R})$ to $L^2(\mathbf{R})$. As both these spaces are Hilbert spaces, I can consider the adjoint $T^*: L^2(\mathbf{R}) \to H^1(\mathbf{R})$. How does this operator relate to the formal adjoint of $T$, i.e. the operator $T^*_F f = -Df$, which satisfies $$ \int Tf \overline{g} = \int f \overline{T^*_F g} $$ for all suitably smooth $f,g$, i.e. $f,g \in C_c^\infty(\mathbf{R})$.
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Integration by parts. – eyeballfrog Jul 21 '22 at 14:31
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Check $T T^*$ argument with which we easily can show if the operator is bounded or not, if it was not mentioned. – Mr. Proof Jul 21 '22 at 14:37
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https://math.stackexchange.com/questions/314113/dual-space-of-h1 – Disintegrating By Parts Aug 01 '22 at 09:53