Let $H$ be a separable Hilbert space and $V$ be an $m$-dimensional proper subspace of $H.$ Let $T \in \mathcal L (H)$ be a compact operator. Show that there exists $0 \neq v_0 \in H$ with $v_0 \perp V$ such that the supremum $$\sup\limits_{v \neq 0,\ v \perp V} \frac {\left \langle Tv, v \right \rangle} {\|v\|^2} = \frac {\left \langle Tv_0, v_0 \right \rangle} {\left \|v_0 \right \|^2}$$ i.e. the supremum is attained.
I tried by taking a maximizing sequence $\{u_n \}_{n \geq 1}$ in $H$ with $\|u_n\| = 1$ and $u_n \perp V$ for all $n \geq 1$ such that the sequence $\{\left \langle T u_n , u_n \right \rangle \}_{n \geq 1}$ converges to the supremum. Since $T$ is compact, the sequence $\{T u_n \}_{n \geq 1}$ will then have a convergent subsequence $\left \{T u_{n_k} \right \}_{k \geq 1}$ converging to some $y \in H.$ But how do I proceed from here? Could anyone give me some suggestion regarding this one?
Thanks a bunch!
EDIT $:$
Let $\alpha = \sup\limits_{v \neq 0,\ v \perp V} \frac {\langle Tv, v \rangle} {\|v\|^2}.$ Choose a sequence $\{u_n \}_{n \geq 1}$ of unit vectors in $V^{\perp}$ such that $\langle T u_n, u_n \rangle \to \alpha.$ Since $T$ is compact, so is $T^{\ast}$ and hence $\left \{T^{\ast} (u_n)\right \}_{n \geq 1}$ has a subsequence $\left \{T^{\ast} (u_{n_k}) \right \}_{k \geq 1}$ converging to some $\beta \in H.$ Now since $H$ is reflexive, $\left \{u_{n_k} \right \}_{k \geq 1}$ has a further subsequence which is weakly convergent. WLOG we may assume that $\left \{u_{n_k} \right \}_{k \geq 1}$ is weakly convergent converging to some $v_0 \in H.$ Then we have $$\begin{align*} \tag{1} \left \langle T u_{n_k}, u_{n_k} \right \rangle & = \left \langle T u_{n_k} - T v_0, u_{n_k} \right \rangle + \left \langle T v_0, u_{n_k} \right \rangle \end{align*}$$
Now, $$\begin{align*} \left \langle T u_{n_k} - T v_0, u_{n_k} \right \rangle & = \left \langle u_{n_k} - v_0, T^{\ast} u_{n_k} \right \rangle \\ & = \left \langle u_{n_k} - v_0, T^{\ast} u_{n_k} - \beta \right \rangle + \left \langle u_{n_k} - v_0, \beta \right \rangle \end{align*}$$
Clearly, both the terms on the RHS tend to zero and hence by $(1)$ we have $\left \langle T v_0, u_{n_k} \right \rangle \to \alpha$ as $\left \langle T u_{n_k}, u_{n_k} \right \rangle \to \alpha.$ But since the sequence $\left \{u_{n_k} \right \}_{k \geq 1}$ converges weakly to $v_0,$ it follows that $\left \langle T v_0, v_0 \right \rangle = \alpha.$
Also, $v_0 \in V^{\perp}$ because for any $w \in V$ we have $$\left \langle v_0, w \right \rangle = \lim\limits_{k \to \infty} \left \langle u_{n_k}, v \right \rangle = 0$$ since $u_{n_k} \in V^{\perp}$ for all $k \geq 1.$
But I can't determine why $\left \|v_0 \right \| = 1.$ Clearly $\|v_0\| \leq 1$ as $$\|v_0\| \leq \liminf\limits_{k \to \infty} \left \|u_{n_k} \right \| = 1.$$ So if $v_0 \neq 0,$ set $w_0 = \frac {v_0} {\left \|v_0 \right \|}.$ Then $\left \|w_0 \right \| = 1$ and $w_0 \perp V$ and so we have $$\alpha = \left \langle T v_0, v_0 \right \rangle = \|v_0\|^2 \left \langle T w_0, w_0 \right \rangle \leq \alpha \left \|v_0 \right \|^2.$$ So if $0 \lt \|v_0\| \lt 1$ we would have a contradiction.
So we are done if $v_0 \neq 0.$ If $v_0 = 0$ then $\alpha = 0.$ So for $\alpha \neq 0$ we are done.
How to tackle $\alpha = 0$ case?