Let $A: H\rightarrow H$ be a contiunuous, linear map on the Hilbert space $H$, $y\in H\setminus \left\{ 0 \right\}$ and $\lambda \in \mathbb{K}$. Show, that if the equation $\lambda x - A x=y$ has at least one solution $x=x_0$, then it has solution with the smallest norm, and this solution is nonzero and unique.
I'm guessing that the Riesz representation theorem could be helpful, but I can't see how it works.
The set of all $x$ satisfying $\lambda x+Ax=y$ is closed, convex, and non-empty. The minimal norm solution is the projection of $0$ onto this set, which exists and is uniquely determined in Hilbert spaces.
There are two theorems:
Theorem 1: Let $X$ be an inner product space and $M \neq \emptyset$ a complete convex subset. Then for all $x \in X$, there exists a unique $y \in M$ so that $$\delta = \inf_{z \in M} \|x-z\|=\|x-y\|.$$
Theorem 2: Let $X$ be as before, fix some $x \in X$, and suppose that $M$ is in fact a complete subspace $Y$. Then $z:=x-y$ is orthogonal to $Y$ if $y$ is the closest point to $x$.
References can be found in chapter 3 of Kreyzwig's book.
From here, you need only show that the set of solutions to $\lambda x-Ax=y$ is indeed closed and convex. It is convex since $(\lambda-A)x=y$ is affine. You should check that it is in fact closed (this follows from continuity) and a subspace.
On the other hand, fix $0 \in X$ and let $M=\{x \in X \mid (\lambda-A)x=y\}$. Then there exists some vector $y \in M$, that minimizes this distance, and is unique. In particular, $\inf_{z \in M} \|z-0\|=\inf_{z \in M}\|z\|=\|y\|$.
It's not so hard to see that this vector $y$ is the projection of $0$ onto $M$.