[proofplan] Assume two elements of $V$ solve the same equations with the same data. Their difference belongs to $V$, and linearity of $L$ and $\gamma_0$ makes both residuals vanish. The a priori estimate then forces the $X$-norm of the included difference to be zero, so the difference is zero in $X$. Since the inclusion $V \hookrightarrow X$ is injective, the two original elements of $V$ are equal. [/proofplan] [step:Subtract two solutions and identify the homogeneous problem] Fix $f \in Y$ and $g \in Z$. Let $\iota:V \to X$ denote the continuous injective inclusion map. Let $0_V \in V$, $0_X \in X$, $0_Y \in Y$, and $0_Z \in Z$ denote the zero elements of their respective normed vector spaces. Let $u \in V$ and $v \in V$ be two solutions of the system with this same data, so \begin{align*} Lu=f \end{align*} and \begin{align*} Lv=f \end{align*} and also \begin{align*} \gamma_0 u=g \end{align*} and \begin{align*} \gamma_0 v=g. \end{align*} Define $w \in V$ by \begin{align*} w := u-v. \end{align*} Since $V$ is a [vector space](/page/Vector%20Space), $w \in V$. By linearity of $L:V \to Y$, \begin{align*} Lw = L(u-v)=Lu-Lv=f-f=0_Y. \end{align*} By linearity of $\gamma_0:V \to Z$, \begin{align*} \gamma_0 w=\gamma_0(u-v)=\gamma_0u-\gamma_0v=g-g=0_Z. \end{align*} [guided] We begin by proving uniqueness for fixed data. Fix $f \in Y$ and $g \in Z$. Let $\iota:V \to X$ denote the continuous injective inclusion map, and let $0_V \in V$, $0_X \in X$, $0_Y \in Y$, and $0_Z \in Z$ denote the zero elements of the four normed vector spaces. Suppose that $u \in V$ and $v \in V$ both solve the problem with this same pair of data. Thus \begin{align*} Lu=f \end{align*} and \begin{align*} Lv=f, \end{align*} while \begin{align*} \gamma_0 u=g \end{align*} and \begin{align*} \gamma_0 v=g. \end{align*} The standard uniqueness move is to subtract the two solutions, because the difference should solve the corresponding homogeneous problem. Define $w \in V$ by \begin{align*} w := u-v. \end{align*} This is an element of $V$ because $V$ is a vector space. Since $L:V \to Y$ is linear, it preserves subtraction, and therefore \begin{align*} Lw = L(u-v)=Lu-Lv=f-f=0_Y. \end{align*} Likewise, since $\gamma_0:V \to Z$ is linear, \begin{align*} \gamma_0 w=\gamma_0(u-v)=\gamma_0u-\gamma_0v=g-g=0_Z. \end{align*} Thus the difference $w$ lies in the homogeneous kernel of both operators. [/guided] [/step] [step:Apply the a priori estimate to force the difference to vanish] Apply the assumed estimate to the element $w \in V$. Since $Lw=0_Y$ and $\gamma_0w=0_Z$, we obtain \begin{align*} \|\iota(w)\|_X \le C\left(\|0_Y\|_Y+\|0_Z\|_Z\right). \end{align*} The norm axioms give $\|0_Y\|_Y=0$ and $\|0_Z\|_Z=0$, so \begin{align*} \|\iota(w)\|_X \le 0. \end{align*} Because norms are nonnegative, $\|\iota(w)\|_X=0$. Hence $\iota(w)=0_X$ by definiteness of the norm on $X$. [/step] [step:Use injectivity of the inclusion to conclude equality in $V$] Since $\iota:V \to X$ is injective and $\iota(w)=0_X=\iota(0_V)$, we have $w=0_V$. Recalling that $w=u-v$, this gives \begin{align*} u-v=0_V. \end{align*} Therefore $u=v$ in $V$. Since the choice of $f \in Y$ and $g \in Z$ was arbitrary, the problem has at most one solution in $V$ for every pair of data $(f,g) \in Y \times Z$. [/step]