A proof pertaining to the projector operator Let $H_{1}$ be any subspace of a Hilbert space $H$, and let $H_{2} = H_{1}^{\bot}$ be the orthogonal complement of $H_{1}$, so that an arbitrary element $h \in H$ has a unique representation of the form $h = h_{1} + h_{2}$ where $h_{1} \in H_{1}$ and $h_{2} \in H_{2}$. Let $Ph = h_{1}$. $P$ is a continuous linear operator known as the projection operator. I want to prove that the projection operator is completely continuous if and only if the subspace $H_{1}$ is finite dimensional. 
I have been thinking about this one for awhile now and have no idea. I would really appreciate the help!!! 
 A: Let's agree that a completely continuous operator $T:X\rightarrow Y$ between Banach spaces is one that sends weakly convergent sequences to norm convergent ones.
For the first part, suppose that $H_1$ is finite dimensional. Consider a weakly convergent sequence $(x_n)_n$ in $H$. The projection $P:H \rightarrow H$ can be correstricted to $H_1$ to give a well defined continuous linear operator $P:H \rightarrow H_1$. Now, $P$ is norm-norm continuous and so it's weak-weak continuous and $(P(x_n))_n$ weakly convergent in $H_1$. Because $H_1$ is finite dimensional, norm and weak topologies agree in $H_1$ and then $(P(x_n))_n$ is norm convergent in $H_1$ and so in $H$. This proves that $P$ is completely continuous.
For the second part, suppose that $P$ is completely continuous. As $H$ is a Hilbert space it is also reflexive and thus every completely continuous $T:H\rightarrow H$ is compact. In particular, $P$ is compact. As $P$ is a projection with image on $H_1$, we have that $P$ agrees with the identity over $H_1$. From this we have that the identity of $H_1$ is a compact operator and so $H_1$ must be finite dimensional.
A: Suppose the range of $P$ is infinite dimensional. Then there exists an orthonormal sequence $\{ e_{n} \}_{n=1}^{\infty}$ in the range of $P$. This is a sequence of unit vectors for which $\{ Pe_{n}=e_{n} \}_{n=1}^{\infty}$ has no convergent subsequence.
