A variational inequality satisfied in a Hilbert space

Question

I'm trying to establish the existance of $u \in K \subset V$, a closed convex subset of vector space with an inner product, such that for a fixed $q \in V$: $$(u-q,v-u) \geq 0 \quad \forall v \in K$$ The proof I'm reading proceeds as follows:

Note that the inequality holds iff $\|u-q\| \leq \|v-q\|$. Since $K$ is a closed convex subset, there exists a unique element of $K$ that minimizes $\|v-q\|$, namely $u = P_Kq$ where $P_K$ is the projection of $q$ onto $K$.

This was a bit terse for me. Why does the inequality hold? From what fact does $u = P_Kq$ follow?

Answer 1

There's an error in the first statement of the proof. The variational inequality proves the inequality in norms, but the converse simply doesn't hold. For example, in $\Bbb{R}^2$ under the dot product, take $u = (0, 0)$, $v = (0, 3)$, and $q = (1, 1)$. Then $$\|u - q\|^2 = 2 \le 5 = \|v - q\|^2,$$ but $$\langle u - q, v - u \rangle = (-1, -1) \cdot (0, 3) = -3 < 0.$$ I suggest finding another proof of this inequality.

But, as for your second question, if you can show that $\|u - q\| \le \|v - q\|$ for all $v \in K$ (and where $u$ is assumed to be in $K$), then you've found a point $u \in K$ that is of minimal distance from $q$. This, by definition, makes $u$ the metric projection of $q$ onto $K$.

EDIT: Actually, I have a proof of this inequality that I wrote up on hand:

Theorem: Suppose $X$ is a real Hilbert space, and $C$ is closed, non-empty, and convex. Let $x \in X$ and $z \in C$. Then, $$\langle x - z, c - z \rangle \le 0 \quad \forall \, c \in C$$ if and only if $z = p_C(x)$.

Proof: Suppose $z, c \in C$, with $z \neq c$. Let $f : \mathbb{R} \to \mathbb{R}$ be defined by \begin{align*} f(\lambda) &= \|x - \lambda c - (1 - \lambda)z\|^2 - \|x - z\|^2 \\ &= 2 \lambda \langle x - z, z - c \rangle + \lambda^2 \|z - c\|^2. \end{align*} Note that this is a convex quadratic in $\lambda$.

Suppose that $z = p_C(x)$. When $\lambda \in [0, 1]$, we have $\lambda c + (1 - \lambda)z \in C$, hence $f(\lambda) \ge 0 = f(0)$. The minimum value of $f$ must be achieved at some $\lambda^* \le 0$. Thus, $$0 \ge \lambda^* = \frac{-\langle x - z, z - c \rangle}{\|z - c\|^2} \iff \langle x - z, c - z \rangle \le 0.$$ Note that the final inequality also holds for when $c = z = p_C(x)$.

Conversely, suppose $z \in C$ such that $\langle x - z, c - z \rangle \le 0$ for all $c \in C$. Therefore, when $c \neq z$, $\lambda^* \le 0$. This implies that $f$ is increasing on the interval $[0, 1]$. Hence, $$\|x - c\| = f(1) \ge f(0) = \|x - z\|.$$ As this holds for arbitrary $c \in C$, we have $z = p_C(x)$.

Answer 2

The statement is incorrect. Consider $V = \mathbb R^2$ and $K = \overline{B}_1(0)$, $u=(0,0)$, $v=(\epsilon,1-\epsilon)$ for some small $\epsilon > 0$ and $q = (2,0)$. Then $\|u-q\|\le\|v-q\|$ and $(u-q,v-u) = -2\epsilon < 0$.

Answer 3

I think the proof is missing an important detail: What we actually have is that for $u \in K$ and $q \in V$ the statement $$ (u - q, v - u ) \ge 0 \qquad \forall v \in K$$ is equivalent to $$ \|u - q \| \le \| v - q \| \qquad \forall v \in K.$$

The proof is given in the answer by Theo Bendit.

The second statement is just $u = \operatorname{proj}_K(q)$.