Inner Product Representation of Functional on $H^1_0$ Let $T\colon H^1_0(\mathbb{R})\to \mathbb{R}, \;T(f)=\int f'\phi  \;dx$, where $\phi\in L^2$ is fixed. By Hölder, $|T(f)|\leq\|\phi\|_2\|f'\|_2\leq C \|f\|_{H^1_0}$, i.e. $T$ is continuous. Therefore, by the Riesz representation theorem there should exist $\psi\in H^1_0$, such that $T(f)=<f,\psi>_{H^1_0}=\int f\psi + f'\psi' \; dx \quad \forall f\in H^1_0$.
My problem is that I don't see how this can be possible. On the other hand, I can't prove it is not possible, so I am willing to believe it but would be thankful for any explanatory remarks/good intuition on that. 
 A: What you really have is
$$
            \int f\phi\,dx = \int f\psi\,dx + \int f'\psi'\,dx.
$$
The $\psi$ is determined uniquely by $\phi$. The question comes down to finding a twice absolutely continuous $\psi$ such that $-\psi''+\psi=\phi$ with $\psi,\psi',\psi''\in L^{2}$, which can be done.
A: Things make more sense when you look at the Fourier transform.  Letting
$\widehat{\phi}$ and $\widehat{f}$ be the Fourier transforms of $\phi$ and $f$,
and ignoring dimensional constants
$$
\int f' \phi = \int i\xi \widehat{f} \widehat{\phi}.
$$
On the other hand,
$$
\int f\psi + f'\psi' = \int (1 + |\xi|^2) \widehat{f} \widehat{\psi}.
$$
If we take $\widehat{\psi} = i\xi \widehat{\phi} / (1 + |\xi|^2)$, then these two expressions match.  This formula tells you that this would also work if $f\in H^{-1}(\mathbb{R})$ instead of just $L^2$.   
As for the intuition, this formula makes a little bit of sense if you try to find $\psi$ using integration by parts.  The first step is to observe
$$
\int f' \phi = -\int f\phi'
$$
and this suggests that $\psi = \phi'$ might be a good first guess.  Maybe this explains the multiplication by $i\xi$.  Now that guess for $\psi$ fails because you can't control the $f'\psi'$ term.  Maybe that explains the $(1+|\xi|^2)^{-1}$ term.  It would be interesting to think more about this ...
