# Compact Operator defined by inner product

Let $H$ be a Hilbert space and $y,z \in H$. Define bounded linear operator $Ax=\langle x,y\rangle z$ where $\langle,\rangle$ is inner product. Would you help me to prove that $A$ is compact operator.

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Let $\{x_n\}_n$ be a bounded sequence in $H$. Then the sequence $\{\langle x_n,y \rangle \}_n$ is bounded in $\mathbb{R}$, and therefore relatively compact: assume it converges, up to a subsequence, to some $r \in \mathbb{R}$. Now, $$\left| \langle x_n,y \rangle z - r z \right| \leq \left| \langle x_n,y \rangle -r \right| \|z\| \to 0,$$ and we conclude that $Ax_n$ converges, up to a subsequence, to $r z$.