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Interpret the angle as the directed subspace angle defined by the infimum of projected unit vectors; it is different from the smallest angle between two subspaces. WriteConsider the bounded linear operator $T:W⟶V^{\perp }$ given by $Tw=P_{V^{\perp }}w$. Its adjoint operator, between these two Hilbert spaces, is $T^{\ast }u=P_{W}u$: for $w\in W$ and $u\in V^{\perp }$, the orthogonal projections give $⟨Tw,u⟩=⟨w,u⟩=⟨w,P_{W}u⟩$. The two positive directed subspace angle cosines yieldThe first bound makes $T$ injective and gives a closed range. Explicitly, if $T{w}_j$ converges, then $\Vert w_j-w_k\Vert \le a^{-1}\Vert T{w}_j-T{w}_k\Vert$, so $(w_j)$ is a Cauchy sequence. The closed subspace of a Hilbert space $W$ is complete, and its limit maps to the proposed range limit. The second bound gives $\ker T^{\ast }=\{0\}$. A vector $u\in V^{\perp }$ orthogonal to the range has $T^{\ast }u=0$, so it must be zero. The range is therefore dense as well as closed in $V^{\perp }$, and $T$ is onto. This is the mechanism of invertibility from lower bounds on an operator and its adjoint.

For any $f\in H$, choose the unique $w\in W$ with $Tw=P_{V^{\perp }}f$. Then $P_{V^{\perp }}(f-w)=0$, so $f-w\in V$. Moreover, if $w\in W\cap V$, then $Tw=0$ and hence $w=0$. Every vector has a unique decomposition, andThe direct sum is a topological one as well: the component $w=T^{-1}{P}_{V^{\perp }}f$ depends boundedly on $f$, with operator norm at most $a^{-1}$.

For precision, the quoted equality of the norms of complementary oblique projections needs both summands nonzero. For example, with $H=\mathbb{R}$, $V=H$ and $W=\{0\}$, the oblique projection is $I$, so $\Vert I\Vert =1$ but $\Vert I-I\Vert =0$. The secant function has value one here, so the second equality in the quoted formula fails. With nonzero complementary summands its intended version is valid. The proof above does not use that formula. The angle itself is undefined on a zero source space because it has no unit vectors; expressing the hypotheses as the two lower bounds handles zero spaces without ambiguity.

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a bounded linear functional. The Riesz representation theorem supplies a unique vector $Lv$ such thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. The invertibility from lower bounds on an operator and its adjoint therefore shows that $L$ is invertible and $L^{-1}$ is bounded.

Set $v_f=L^{-1}f$. Then $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus the unique vector is $v_f=L^{-1}f$.