[proofplan] We convert the bilinear form equation $B[u, v] = f(v)$ into an operator equation $Au = g$ on $H$. For each $u \in H$, the map $v \mapsto B[u, v]$ defines a bounded linear functional, and the [Riesz Representation Theorem](/theorems/221) produces a unique element $Au \in H$ representing it. The boundedness hypothesis on $B$ makes $A$ a bounded operator, while coercivity makes $A$ bounded below — giving injectivity, closedness of the range, and (via the orthogonal decomposition) surjectivity. The equation $Au = g$ then has a unique solution. [/proofplan] [step:Construct the operator $A: H \to H$ via the Riesz Representation Theorem] Fix $u \in H$. Define the map \begin{align*} \psi_u: H &\to \mathbb{R} \\ v &\mapsto B[u, v]. \end{align*} By bilinearity of $B$ in the second argument, $\psi_u$ is linear. By the boundedness hypothesis: \begin{align*} |\psi_u(v)| = |B[u, v]| \le \alpha \|u\| \|v\|, \end{align*} so $\psi_u \in H^*$ with $\|\psi_u\|_{H^*} \le \alpha \|u\|$. By the [Riesz Representation Theorem](/theorems/221), there exists a unique element in $H$ representing $\psi_u$. Let $\tau: H^* \overset{\sim}{\to} H$ denote the Riesz isomorphism and define $Au := \tau(\psi_u)$. By construction: \begin{align*} (Au, v)_H = \psi_u(v) = B[u, v] \qquad \text{for all } u, v \in H. \end{align*} The operator $A: H \to H$ is linear, since $B$ is linear in the first argument and $\tau$ is linear. [guided] The goal is to rephrase the problem $B[u, v] = f(v)$ as an operator equation that we can solve directly. The key idea: for each fixed $u$, the map $v \mapsto B[u, v]$ eats vectors in $H$ and produces scalars — it is a linear functional. If this functional is also bounded, the [Riesz Representation Theorem](/theorems/221) tells us it is represented by a unique element of $H$, which we call $Au$. Define the map \begin{align*} \psi_u: H &\to \mathbb{R} \\ v &\mapsto B[u, v]. \end{align*} Linearity of $\psi_u$ follows from bilinearity of $B$ in the second argument. Boundedness follows from the hypothesis $|B[u, v]| \le \alpha \|u\| \|v\|$: \begin{align*} |\psi_u(v)| = |B[u, v]| \le \alpha \|u\| \|v\|, \end{align*} so $\psi_u \in H^*$ with $\|\psi_u\|_{H^*} \le \alpha \|u\|$. Now the [Riesz Representation Theorem](/theorems/221) applies: since $H$ is a [Hilbert space](/pages/1083) and $\psi_u \in H^*$, there exists a unique $w \in H$ with $\psi_u(v) = (w, v)_H$ for all $v \in H$. Let $\tau: H^* \overset{\sim}{\to} H$ denote the Riesz isomorphism, and define $Au := \tau(\psi_u) = w$. This gives the fundamental identity: \begin{align*} (Au, v)_H = \psi_u(v) = B[u, v] \qquad \text{for all } u, v \in H. \end{align*} The operator $A: H \to H$ is linear because $B$ is linear in $u$ (fixing $v$) and $\tau$ is linear. [/guided] [/step] [step:Show that boundedness of $B$ implies $\|Au\| \le \alpha \|u\|$] Since $\tau$ is an isometry, we have: \begin{align*} \|Au\| = \|\tau(\psi_u)\| = \|\psi_u\|_{H^*} \le \alpha \|u\|. \end{align*} Hence $A \in \mathcal{L}(H)$ with $\|A\|_{\mathcal{L}(H)} \le \alpha$. [/step] [step:Show that coercivity of $B$ implies $\|Au\| \ge \beta \|u\|$] Substituting $v = u$ in the fundamental identity and applying the Cauchy–Schwarz inequality: \begin{align*} \beta \|u\|^2 \le B[u, u] = (Au, u)_H \le \|Au\| \|u\|. \end{align*} For $u \ne 0$, dividing both sides by $\|u\|$ gives: \begin{align*} \|Au\| \ge \beta \|u\|. \end{align*} The case $u = 0$ holds as $0 \ge 0$. This lower bound immediately gives **injectivity**: if $Au = 0$, then $\beta \|u\| \le 0$, so $u = 0$. [guided] Why do we care about a lower bound on $\|Au\|$? Because $\|Au\| \ge \beta \|u\|$ gives us two things at once: injectivity (which is immediate) and closedness of the range (which we prove in the next step). Together, these will force $A$ to be surjective. Substituting $v = u$ in the identity $(Au, v)_H = B[u, v]$ and applying the Cauchy–Schwarz inequality on the inner product $(Au, u)_H$: \begin{align*} \beta \|u\|^2 \le B[u, u] = (Au, u)_H \le \|Au\| \|u\|. \end{align*} For $u \ne 0$, dividing both sides by $\|u\|$: \begin{align*} \|Au\| \ge \beta \|u\|. \end{align*} The case $u = 0$ is immediate. For injectivity: if $Au = 0$, then $\beta \|u\| \le \|Au\| = 0$, so $\|u\| = 0$ and $u = 0$. [/guided] [/step] [step:Prove that $\operatorname{Range}(A)$ is closed in $H$] Let $(y_k)_{k=1}^\infty \subset \operatorname{Range}(A)$ with $y_k \to y$ in $H$. Write $y_k = Au_k$. By the lower bound: \begin{align*} \|u_k - u_m\| \le \frac{1}{\beta}\|Au_k - Au_m\| = \frac{1}{\beta}\|y_k - y_m\|. \end{align*} Since $(y_k)$ is Cauchy, so is $(u_k)$. By completeness of $H$ (see [Complete Metric Space](/pages/1230)), there exists $u \in H$ with $u_k \to u$. Since $A$ is bounded (hence [continuous](/pages/1200)): \begin{align*} y = \lim_{k \to \infty} y_k = \lim_{k \to \infty} Au_k = Au \in \operatorname{Range}(A). \end{align*} [/step] [step:Prove that $A$ is surjective via orthogonal decomposition] Suppose for contradiction that $\operatorname{Range}(A) \ne H$. Since $\operatorname{Range}(A)$ is a closed subspace of $H$, the [orthogonal decomposition theorem](/theorems/241) gives a non-zero element $z \in \operatorname{Range}(A)^\perp$. In particular, $Az \in \operatorname{Range}(A)$, so $(Az, z)_H = 0$. But then: \begin{align*} 0 = (Az, z)_H = B[z, z] \ge \beta \|z\|^2 > 0, \end{align*} a contradiction. Therefore $\operatorname{Range}(A) = H$. [guided] We have shown that $A$ is injective and has closed range. Why does this force surjectivity? In a [Hilbert space](/pages/1083), every closed subspace $M$ satisfies $H = M \oplus M^\perp$. If $\operatorname{Range}(A) \ne H$, then $\operatorname{Range}(A)^\perp$ contains a non-zero element $z$. Since $z \in H$, we can compute $Az \in \operatorname{Range}(A)$. But $z \in \operatorname{Range}(A)^\perp$, so $(Az, z)_H = 0$. Now the coercivity hypothesis produces a contradiction: \begin{align*} 0 = (Az, z)_H = B[z, z] \ge \beta \|z\|^2 > 0. \end{align*} This contradiction shows $\operatorname{Range}(A)^\perp = \{0\}$, and therefore $\operatorname{Range}(A) = H$. [/guided] [/step] [step:Solve $Au = g$ and conclude existence and uniqueness] Given $f \in H^*$, the [Riesz Representation Theorem](/theorems/221) provides a unique $g \in H$ with $f(v) = (g, v)_H$ for all $v \in H$. Since $A: H \to H$ is a bijection, there exists a unique $u \in H$ with $Au = g$. Then for every $v \in H$: \begin{align*} B[u, v] = (Au, v)_H = (g, v)_H = f(v). \end{align*} Uniqueness follows from injectivity of $A$: if $B[u_1, v] = B[u_2, v] = f(v)$ for all $v$, then $(A(u_1 - u_2), v)_H = 0$ for all $v$, so $A(u_1 - u_2) = 0$, and injectivity gives $u_1 = u_2$. [/step]