suppose $H$ is a hilbert space and $a,b \in H$, where $(a,b)>0$. Prove there exists a unique $x \in H$ of minimal norm such that: suppose $H$ is a hilbert space and $a,b \in H$, where $(a,b)>0$. Prove there exists a unique $x \in H$ of minimal norm such that:
$(x,a) \geq 1$ and $(x,b) \geq 1$
I thought of using something like Riesz lemma, but i couldn't quite get the result partly because this lemma is mainly used for norms. Any hints,tips or tricks i can use for this question.
PS:
The only thing i know about Hilbert spaces is that they are complete with respect to the metric associated with the norm induced by the inner product.
 A: A standard Hilbert space result is that any non empty closed convex set has a
unique point of minimum norm.
Let $C= \{ x | (x,a) \ge 1, (x,b) \ge 1 \}$. This is a closed convex set. It
is easy to verify that $\max({1 \over \|a\|^2}, {1 \over (a,b)}) a \in C$,
so the set is non empty.
The proof starts by taking a sequence of elements $x_n \in C$ such that $\|x_n\| \to \inf_{x \in C} \|x\|$ and showing that the sequence in Cauchy.
This establishes existence. Uniqueness follows because the norm is strictly convex.
A: Based on your background, I won't assume any advanced knowledge.
One way to solve is to reduce the problem to a problem in $\mathbb{R}^2$, where I assume you can show what you want. To see how the reduction works, suppose $\{ a,b\}$ is a linearly-independent set of vectors. Then every $x\in H$ can be uniquely represented as $x=\alpha a+ \beta b + y$ where $y\perp a$ and $y\perp b$. Then
$$
         (x,a) = (\alpha a + \beta b,a) \\
         (x,b) = (\alpha a + \beta b,b) \\
         \|x\|^2=\|\alpha a+\beta b\|^2+\|y\|^2.
$$
Because you want to minimize $\|x\|$ subject to $(x,a)\ge 1$ and $(x,b)\ge 1$, it is clear that any such solution $x_0$ must have the form $\alpha a+\beta b$.
The problem in $\mathbb{R}^2$ can be written as a minimization problem in two real variables $\alpha$, $\beta$ involving the fixed numbers $(a,a)$, $(a,b)=(b,a)$, $(b,b)$. The new problem is to minimize
$$
           f(\alpha,\beta)= \|\alpha a+\beta b\|^2 = \alpha^2(a,a)+2\alpha\beta(a,b)+\beta^2(b,b)
$$
subject to the constraints:
$$
        1 \le \alpha(a,a)+\beta(b,a) = (\alpha\hat{i}+\beta\hat{j})\cdot((a,a)\hat{i}+(b,a)\hat{j}) \\
        1 \le \alpha(a,b)+\beta(b,b) = (\alpha\hat{i}+\beta\hat{j})\cdot((a,b)\hat{i}+(b,b)\hat{j})
$$
Each constraint equation describes a closed half-plane in $\mathbb{R}^2$ that does not contain the origin.
Start with $f(\alpha,\beta)=r$, which is the equation of an ellipse, and increase $r$ until the ellipse intersects both half planes described above.
