# A variational inequality satisfied in a Hilbert space

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?

• Puh, what is $q$? – amsmath Mar 24 at 18:52
• my bad $q \in V$ – yoshi Mar 24 at 19:03
• You should note that $V$ has to be complete, otherwise the projection might not exist. – gerw Mar 25 at 7:10

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 $$z = p_C(x)$$.

• What is $K$ in your example? – amsmath Mar 24 at 19:27
• $K$ is the closed, non-empty, convex set onto which we are projecting, as in the question. – Theo Bendit Mar 24 at 19:28
• You should specify it in your example. Otherwise it is incomplete. – amsmath Mar 24 at 19:29
• @amsmath Actually, in my counterexample, there is no such $K$ (I thought you were referring to the second paragraph). My point was that the two inequalities were not equivalent; in particular, if the OP was trying to manipulate one inequality into the other, they were not going to succeed. – Theo Bendit Mar 24 at 20:03
• Of course they are not equivalent. But the points $u$ and $v$ are supposed to be in a closed convex set. However, in your example you can just choose the segment from $u$ to $v$ as $K$. Since one always can do that, your counterexample is actually valid without specifying $K$. – amsmath Mar 24 at 21:02

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 $$(u-q,v-u) = -2\epsilon < 0$$.

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 $$u = \operatorname{proj}_K(q)$$.