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\begin{document}
\header{9.520}{1}
\problem{3}

Finding the minimum of $H[f]$ is a ``simple'' case of taking the
derivative and setting it to 0.  The difficulty arises by the fact that we are differentiating with respect to a function, $f$.
\par
First, let's rewrite $H[f]$ as

\begin{equation}
H[f] = V[f] + \lambda \phi(f)
\end{equation}

so, we want to solve for f in

\begin{equation}
\frac{\delta H}{\delta f(y)} = \frac{\delta V[f]}{\delta f(y)} + \lambda \frac{\delta \phi[f]}{\delta f(y)} = 0
\label{eqz}
\end{equation}

Fist solving for $\frac{\delta V[f]}{\delta f(y)}$:

\begin{equation}
\frac{\delta V[f]}{\delta f(y)} = \int dx 2\left[\int dt \theta(x-t)f(t) - F_l(x)\right]\int dt \theta(x-t)\frac{\delta f(t)}{\delta f(y)}
\end{equation}

which simplifies to:

\begin{equation}
\frac{\delta V[f]}{\delta f(y)} = \int dx 2\left[\int dt \theta(x-t)f(t) - F_l(x)\right] \theta(x-y) = 2(\theta \ast f \ast \theta)(y) - (F_l \ast \theta)(y)
\end{equation}

Taking the Fourier transform,

\begin{equation}
{\cal F}\left[\frac{\delta V[f]}{\delta f(y)}\right] = 2\tilde{\theta}(\omega)\tilde{f}(\omega) - \tilde{F}_l\tilde{\theta}
\end{equation}

Given that $\tilde{\theta} = \frac{1}{\omega}$ and $\tilde{F}_l =
\frac{1}{l}\sum^{l}_{i=1} \tilde{\theta} e^{i\omega t_i}$, we get the
final expression:

\begin{equation}
{\cal F}\left[\frac{\delta V[f]}{\delta f(y)}\right] = \frac{2}{\omega^2}\tilde{f}(\omega) - \frac{1}{\omega^2 l} \sum^{l}_{i=1} e^{i\omega t_i}
\label{p1}
\end{equation}

Now, we use the general case of $\phi[f]$, we obtain:

\begin{equation}
\frac{\delta \phi[f]}{\delta f(y)} = \frac{\delta}{\delta f(y)} \int d\omega \frac{|\tilde{f}(\omega)|^2}{\tilde{G}(\omega)}
\end{equation}

Since $|\tilde{f}(\omega)|^2 =
\tilde{f}(\omega)\tilde{f}^\ast(\omega)$, we can rewrite as:

\begin{equation}
 = \frac{\delta}{\delta f(y)} \int d\omega \frac{1}{\tilde{G}(\omega)} \int dx e^{i\omega x} f(x) \int dt e^{-i\omega t} f(t)
\end{equation}

Bringing all the terms that are constant with respect to f(y) out front:
\begin{equation}
=  \int d\omega dx dt \frac{e^{i \omega (x-t)}}{\tilde{G}(\omega)} \frac{\delta}{\delta f(y)} f(x) f(t)
\end{equation}

which simplifies to 

\begin{equation}
= 2 \int d\omega dt \frac{e^{i \omega (y-t)} f(t)}{\tilde{G}(\omega)}
\end{equation}

which makes

\begin{equation}
{\cal F}\left[\frac{\delta \phi[f]}{\delta f(y)}\right] = \frac{2\tilde{f}(\omega)}{\tilde{G}(\omega)}
\label{p2}
\end{equation}

Combining equations \ref{eqz}, \ref{p1} and \ref{p2}:

\begin{equation}
\frac{\tilde{f}(\omega)}{\omega^2} - \frac{1}{\omega^2}\frac{1}{l} \sum_{i=1}^{l} e^{i \omega t_i} + \lambda \frac{\tilde{f}(\omega)}{\tilde{G}(\omega)} = 0
\end{equation}

Therefore,

\begin{equation}
\tilde{f}(\omega) = \frac{1}{1+\lambda \frac{\omega^2}{\tilde{G}(\omega)}} \frac{1}{l} \sum_{i=1}^{l} e^{i\omega t_i}
\end{equation}

If we define the part before the $\frac{1}{l}$ to be the Fourier
transform of $\Phi(x)$,

\begin{equation}
f(x) = \frac{1}{l} \sum_{i=1}^{l} \Phi[x-t_i]
\end{equation}

which is Parzen windows, with a kernel function dependent on the
regularization function $\phi$.
\end{document}