In regards to the ill-posedness of the problem of differentiation of noisy data, that is, to find the derivative of a function $f_\delta(x) = f(x) + \sqrt{2}\delta\sin(2\pi kx)$, I am told that I can regularize the noisy data $f_\delta$ by the Tikhonov regularization.
This technique consists in the minimizing of the function
\begin{equation} J(u) := \frac{1}{2} \int_0^1 (u(x) - f_\delta(x))^2 \ dx + \frac{\alpha}{2} \int_0^1 |u'(x)|^2 \ dx \end{equation} over all functions $u$ in the Sobolev space $H^1([0,1]) := \{u \in L^2([0,1]) : u' \in L^2([0,1])\}$, such that $u(0)=u(1)=0$.
Then, I need to prove that the functional $J$ has a unique minimum, which is achieved at the function $u_\alpha$, a solution to the problem
\begin{cases} u_\alpha(x) - \alpha \cdot u''_\alpha(x) = f_\delta(x) & x \in (0,1), \\ u_\alpha(0) = u_\alpha(1) = 0. & \end{cases}
So, my approach was to differentiate the function $J$ with respect to $u$. My attempt was
\begin{equation} \frac{d(J(u))}{du} := \frac{1}{2} \int_0^1 \frac{d((u(x) - f_\delta(x))^2)}{du} \ dx + \frac{\alpha}{2} \int_0^1 \frac{d((u'(x))^2)}{du} \ dx = \int_0^1 u(x) - f_\delta(x) \ dx + \frac{\alpha}{2} \int_0^1 \frac{d((u'(x))^2)}{du} \ dx \end{equation}
and I'm not really sure on how to continue from here. I can see that I am close, but there's something wrong in the process that I just don't seem to understand.
Any help would be appreciated.
Using integration by part on $u'(x)^2 = u'(x)u'(x)$, we obtain,
\begin{equation} \int_0^1 u'(x)^2\mathrm{d}x = \left[u(x)u'(x)\right]_0^1 - \int_0^1 u(x)u''(x)\mathrm{d}x. \end{equation}
Using the boundary conditions on $u_\alpha$ the first term on the right-hand side vanishes, so we have, $$\frac{\mathrm{d}}{\mathrm{d}u}\int_0^1 u_\alpha'(x)^2\mathrm{d}x = - \int_0^1 \frac{\mathrm{d}}{\mathrm{d}u}u_\alpha(x)u_\alpha''(x)\mathrm{d}x = - \int_0^1 u_\alpha''(x)\mathrm{d}x,$$ hence $u_\alpha$ is a solution to $\frac{\mathrm{d}J}{\mathrm{d}u}(u)=0$.
So far what is missing is to justify that $J$ has only a unique critical point (most likely through a strict convexity argument).