How can we say that solution of following functional exist under some conditons?

Question

If $F: X \to Y$ is continuous non linear operator satisfying $F(x) = y$ where $X, Y$ are hilbert spaces. If $F$ is continuous and weakly sequentially closed, can we say that the functional $$||F(x)-y^{\delta}||^2+\alpha||x-x^*||^2 \to \text{min}, \quad x\in D(F)$$ has a solution? it is unique or not? Here $y^{\delta}$ is approximation of $y$, and $\alpha > 0, x^* \in X$.

Answer 1

Let $f : X \to [0, \infty)$ be given by $f(u) = \| F(x) - y^\delta \|^2 + \alpha \| x - u \|^2$. Since $f$ is bounded below by $0$, there exist $i = \inf f$ a finite number. This means that there exist a sequence $(u_n) \subset X$ such that $f(u_n) \to i$. We want to show that $(u_n)$ has at least one limit point $x^*$ such that $f(x^*) = i$ (and thus $x^*$ is a desired solution to the minimization problem, in general not unique).

Writing the definition of convergence "with $\epsilon$ and $N_\epsilon$" for $f(u_n) \to i$, choose $\epsilon = 1$ to get an $N$ such that for $n \ge N$ we have $| f(u_n) - i | < 1$, which implies $f(u_n) < i + 1$. Explicitly,

$$\| F(x) - y^\delta \|^2 + \alpha \| x - u_n \|^2 < i + 1 ,$$

whence it follows (by neglecting the first square) that

$$\| x - u_n \| \le \sqrt {\frac {i+1} \alpha} ,$$

which means that for $n \ge N$ all the $u_n$ live inside the ball of center $x$ and radius $\sqrt {\frac {i+1} \alpha}$. It is an easy consequence of Alaoglu's theorem (and of the fact that every Hilbert space is isomorphic to its dual) that closed balls in a Hilbert space are weakly compact. This means that for $n \ge N$ all the $u_n$ live inside a weakly compact subset, therefore there must exist a weakly convergent subsequence of $(u_n)_{n \ge N}$. Let this subsequence be $(x_n)$ and its limit be $x^*$. We want to show that $f(x^*) = i$.

Consider the sequence of pairs $(x_n, f(x_n)) \in X \times [0, \infty)$. This sequence has the limit $(x^*, i) \in X \times [0, \infty)$. The fact that $F$ is weakly sequentially closed implies that the graph of $F$ is weakly sequentially closed in $X \times Y$, which in turn implies that the graph of $f$ is weakly sequentially closed in $X \times [0, \infty)$. Since the pairs $(x_n, f(x_n))$ all belong to this (weakly sequentially closed) graph, it follows that the limit of the sequence $(x_n, f(x_n))$ must belong to the graph, i.e. $(x^*, i)$ belongs to the graph, i.e. $f(x^*) = i$.

Notice that, since there may be several weakly convergent subsequences of $(u_n)$, there may in principle be several limit points $x^*$, therefore the solution of your problem need not be unique, in general.

Notice also that we haven't really used anything about $y^\delta$, the fact that it is an approximation of $y$ plays no role.