Coercivity of self-adjoint operator

Question

Let $X$, $Y$ be real Hilbert spaces. Let $A\in\mathcal{L}(X,Y)$, let $A^{\ast}$ denote the Hilbert-adjoint, and let $I_X$, $I_Y$ denote the identity mappings on $X$, $Y$. Given the self-adjoint operator $B:X\times Y\to X\times Y$, \begin{equation*} B=\begin{pmatrix}A^{\ast}A+I_X & A^{\ast}\\ A & I_Y\end{pmatrix}, \end{equation*} how can I show that $B$ is coercive, i.e., there exists $c>0$ such that $\big\langle B(h_1,h_2),(h_1,h_2)\big\rangle_{X\times Y}\geq c\lVert(h_1,h_2)\rVert_{X\times Y}^2$ for all $(h_1,h_2)\in X\times Y$? Clearly, in the case $X=\mathbb{R}^{n}$, $Y=\mathbb{R}^m$, the Schur complement answers the question (with the smallest eigenvalue of $B$).

Answer 1

Inserting $h=(h_1,h_2)$ we obtain $$ (Bh,h) = \|Ah_1\|_Y^2 + \|h_1\|_X^2 + \|h_2\|_Y^2 + 2(h_2, Ah_1)_Y = \|h_1\|_X^2 + \|Ah_1+h_2\|_Y^2. $$ Assume that $B$ is not coercive, i.e., for every $k$ there are $h_{1,k}\in X$ and $h_{2,k}\in Y$ with $\|h_{1,k}\|_X^2 + \|h_{2,k}\|_Y^2 = 1$ and $$ \|h_{1,k}\|_X^2 + \|Ah_{1,k}+h_{2,k}\|_Y^2 = (Bh_k,h_k) < 1/k. $$ It follows $h_{1,k}\to0$ in $X$ and $Ah_{1,k}+h_{2,k}\to0$ in $Y$. This implies $h_{2,k}\to0$ in $Y$, which is a contradiction to $\|h_{1,k}\|_X^2 + \|h_{2,k}\|^2 = 1$.

Surprisingly, no compactness argument is needed.

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Here is a much simpler approach: The operator $B$ can be written as $$ B=\pmatrix{I & A^*\\0&I}\pmatrix{I & 0\\A&I} =: L^*L. $$ The factor $L$ is invertible, $L^{-1}=\pmatrix{I & 0\\-A&I}$. Then $$ (Bx,x) = \|Lx\|^2 \ge \|L^{-1} \|^{-2} \cdot \|x\|^2, $$ where I used $\|L^{-1}v\|\le \|L^{-1}\|\cdot \|v\|$ with $v=Lx$.