Consider the Hilbert space $L^2([0,1])$ (with Lebesgue measure $dm$). Let $\eta \in L^2([0,1])$ be strictly positive almost everywhere, and essentially bounded, and consider the (nonlinear) functional $\phi: L^2([0,1]) \rightarrow \mathbb{R}$ given by $\phi(f) = \int f^2 \eta \: dm$.
I would like help solving the following optimization problem. For a fixed $g \in L^2([0,1])$ with $\phi(g) = 1$, and some $r \geq 0$, I would like to find the point $g’ \in L^2([0,1])$ which is $L^2$-closest to $g$ and satisfies $\phi(g’) = r$.
My intuition is telling me that $g’$ should be some scaling of $g$ involving $r$ and $\sqrt{f}$. The only potentially helpful thing I’ve noticed is that we may define different Hilbert space structure on $L^2([0,1])$ with inner product $\langle \alpha, \beta \rangle_{\eta} = \int \alpha \beta \eta \: dm$. Denote this Hilbert space by $L^2_\eta ([0,1])$. It is clear that $\{h : \phi(h) = r\}$ defines the sphere of radius $\sqrt{r}$ centered at $0$ in $L^2_{\eta}([0,1])$. Further, the map $f \mapsto \sqrt{\eta} f$ is a unitary mapping from $L^2_{\eta}([0,1])$ to $L^2([0,1])$.
However, I haven’t been able to make this be of use. The issue is that, I’m trying to minimize distance to to the sphere in $L^2_\eta([0,1])$ with respect to the $L^2$ distance instead of the $L^2_\eta$ distance.
Any help would be greatly appreciated. Thanks!
This should be regarded as a long comment more than an actual answer, as I myself am not fully confident in all the details. The idea is a combination of Lagrange multiplier and Euler-Lagrange equation. Namely, since we are trying to minimize the square of the $L^2$-distance $\|f - g\|^2 = \int_0^1 (f - g)^2 \, dm$ subject to the constraint $\phi(f) = \int_0^1 f^2 \eta \, dm = r$, we introduce the Lagrange multiplier $\lambda$ and set $F_\lambda(f) = \|f - g\|^2 + \lambda (\phi(f) - r) = \int_0^1 [(f - g)^2 + \lambda f^2 \eta - \lambda r] \, dm$. Assuming some "sufficient smoothness", a stationary point of $F_\lambda$ should be a pair $(g', \lambda)$ subject to both the constraint $\int_0^1 g'^2 \eta \, dm = r$ as well as the Euler-Lagrange equation for $F_\lambda$: $2g' - 2g + 2\lambda\eta g' = 0$, which yields the solution:
$$g' = \frac{g}{1 + \lambda\eta} \,\mathrm{where}\,\lambda\,\mathrm{satisfies}\int_0^1 \frac{\eta g^2}{(1+\lambda\eta)^2} \, dm = r$$
There is some reason to find this believable, at least when $0 < r \leq 1$. Indeed, in that case it is clear that the $g'$ that is $L^2$-closest to $g$ subject to $\phi(g') = r$ is the same as the $g'$ that is $L^2$-closest to $g$ subject to $\phi(g') \leq r$. As $\{f: \phi(f) \leq r\}$ is closed and convex, such a $g'$ exists and is unique. And the same holds for $\lambda$ with $\int_0^1 \frac{\eta g^2}{(1+\lambda\eta)^2} \, dm = r$. Indeed, when $\lambda = 0$ we have $\int_0^1 \frac{\eta g^2}{(1+\lambda\eta)^2} \, dm = \int_0^1 \eta g^2 \, dm = 1 \geq r$. Because $\eta$ is strictly positive, the function $\mathbb{R}_+ \ni \lambda \mapsto \int_0^1 \frac{\eta g^2}{(1+\lambda\eta)^2} \, dm$ is easily seen to be strictly decreasing. It is also continuous and converges to $0$ as $\lambda \to \infty$, both of which can be proved via dominated convergence theorem. So $\lambda \geq 0$ with $\int_0^1 \frac{\eta g^2}{(1+\lambda\eta)^2} \, dm = r$ exists and is unique as well.
Of course, the above isn't really a rigorous justification, given that I'm not sure what "sufficient smoothness" entails in this situation, as I've never really applied Lagrange multiplier to the optimization on an infinite-dimensional space before. If there's a problem with this reasoning, please feel free to point out.