[proofplan] We show that any two nonzero elements of the real first Dirichlet eigenspace are scalar multiples of one another. The positivity hypothesis in the theorem statement allows us to choose their signs so that both are nonnegative $\mathcal{L}^n$-a.e. and positive on sets of positive $\mathcal{L}^n$-measure. We then subtract a carefully chosen multiple whose integral agrees with the first eigenfunction; the difference is still a first eigenfunction but has zero integral. Positivity forces any nonzero first eigenfunction to have one sign, which is incompatible with having integral zero, so the difference must vanish. [/proofplan]

[step:Choose positive representatives of two first eigenfunctions]Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and let $\mathcal{B}(U)$ denote the Borel $\sigma$-algebra of $U$. For $1 \leq p < \infty$, let $L^p(U)$ denote $L^p(U, \mathcal{B}(U), \mathcal{L}^n)$. Let $C_c^\infty(U)$ denote the real [vector space](/page/Vector%20Space) of smooth functions $\psi: U \to \mathbb{R}$ with compact support in $U$. Let $H^1(U)$ denote the real [Sobolev space](/page/Sobolev%20Space) of functions in $L^2(U)$ whose first weak partial derivatives belong to $L^2(U)$. Let $H_0^1(U)$ denote the closure of $C_c^\infty(U)$ in $H^1(U)$. Let $E_{\lambda_1}$ denote the real first Dirichlet eigenspace in $H_0^1(U)$, that is, the real vector space of weak solutions $\phi \in H_0^1(U) \setminus \{0\}$ of the eigenvalue equation with eigenvalue $\lambda_1$, together with the zero function. Let $u, v \in E_{\lambda_1}$ be nonzero real-valued first eigenfunctions. We use the positivity hypothesis in the statement: every nonzero real first Dirichlet eigenfunction has a fixed sign $\mathcal{L}^n$-a.e. in $U$, and after choosing the nonnegative sign it is positive on a set of positive $\mathcal{L}^n$-measure. Applying this hypothesis to $u$ and $v$ is valid because $u$ and $v$ are nonzero elements of $E_{\lambda_1}$. Replacing $u$ by $-u$ if necessary and replacing $v$ by $-v$ if necessary, we may assume \begin{align*} u(x) \geq 0 \text{ for } \mathcal{L}^n\text{-a.e. } x \in U \end{align*} and \begin{align*} v(x) \geq 0 \text{ for } \mathcal{L}^n\text{-a.e. } x \in U. \end{align*} Since $u$ and $v$ are nonzero as elements of $L^2(U)$, each is positive on a set of positive $\mathcal{L}^n$-measure after this choice of sign. Since the hypotheses on $U$ include boundedness, $\mathcal{L}^n(U) < \infty$. Since $u, v \in H_0^1(U) \subset L^2(U)$, the [Cauchy-Schwarz Inequality](/theorems/432) applied to $|u| \cdot 1$ and $|v| \cdot 1$ on $(U,\mathcal{B}(U),\mathcal{L}^n)$ gives $u, v \in L^1(U)$. Moreover, \begin{align*} \int_U v \, d\mathcal{L}^n(x) > 0. \end{align*} Thus the scalar \begin{align*} a := \frac{\int_U u \, d\mathcal{L}^n(x)}{\int_U v \, d\mathcal{L}^n(x)} \end{align*} is well-defined and satisfies $a > 0$.[/step]

[guided]Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let $\mathcal{B}(U)$ denote the Borel $\sigma$-algebra of $U$. For $1 \leq p < \infty$, let $L^p(U)$ denote $L^p(U, \mathcal{B}(U), \mathcal{L}^n)$. Let $C_c^\infty(U)$ denote the real vector space of smooth functions $\psi: U \to \mathbb{R}$ with compact support in $U$. Let $H^1(U)$ denote the real Sobolev space of functions in $L^2(U)$ whose first weak partial derivatives belong to $L^2(U)$. Let $H_0^1(U)$ denote the closure of $C_c^\infty(U)$ in $H^1(U)$. Let $E_{\lambda_1}$ denote the real first Dirichlet eigenspace in $H_0^1(U)$, that is, the real vector space of weak solutions $\phi \in H_0^1(U) \setminus \{0\}$ of the eigenvalue equation with eigenvalue $\lambda_1$, together with the zero function. Let $u, v \in E_{\lambda_1}$ be nonzero real-valued eigenfunctions. The key input is the positivity hypothesis in the statement. It applies to both $u$ and $v$ because they are nonzero elements of $E_{\lambda_1}$. Its content is that every nonzero real first Dirichlet eigenfunction has a fixed sign $\mathcal{L}^n$-a.e. in $U$, and after choosing the nonnegative sign it is positive on a set of positive $\mathcal{L}^n$-measure. Therefore, after replacing an eigenfunction by its negative if needed, we may arrange that \begin{align*} u(x) \geq 0 \text{ for } \mathcal{L}^n\text{-a.e. } x \in U \end{align*} and \begin{align*} v(x) \geq 0 \text{ for } \mathcal{L}^n\text{-a.e. } x \in U. \end{align*} Since $u$ and $v$ are nonzero as elements of $L^2(U)$, each is positive on a set of positive $\mathcal{L}^n$-measure after this choice of sign. We want to subtract a multiple of $v$ from $u$ so that the result has zero integral. This requires the integrals of $u$ and $v$ to be finite and the integral of $v$ to be nonzero. Since the hypotheses on $U$ include boundedness, $\mathcal{L}^n(U) < \infty$. Since $u, v \in H_0^1(U)$, in particular $u, v \in L^2(U)$. Applying the [Cauchy-Schwarz Inequality](/theorems/432) to $|u| \cdot 1$ and $|v| \cdot 1$ on the [measure space](/page/Measure%20Space) $(U,\mathcal{B}(U),\mathcal{L}^n)$ gives $u, v \in L^1(U)$. Because $v \geq 0$ $\mathcal{L}^n$-a.e., $v$ is positive on a set of positive $\mathcal{L}^n$-measure, and $v \in L^1(U)$, its integral is strictly positive: \begin{align*} \int_U v \, d\mathcal{L}^n(x) > 0. \end{align*} Hence we may define \begin{align*} a := \frac{\int_U u \, d\mathcal{L}^n(x)}{\int_U v \, d\mathcal{L}^n(x)}. \end{align*} The numerator and denominator are both positive, so $a > 0$.[/guided]

[step:Subtract the integral-matching multiple] Define $w \in H_0^1(U)$ as the equivalence class \begin{align*} w := u - a v. \end{align*} Equivalently, after choosing measurable representatives of $u$ and $v$, a representative of $w$ is given by $x \mapsto u(x) - a v(x)$ for $\mathcal{L}^n$-a.e. $x \in U$. Since $E_{\lambda_1}$ is a real vector space and $u, v \in E_{\lambda_1}$, we have $w \in E_{\lambda_1}$. By the definition of $a$, \begin{align*} \int_U w \, d\mathcal{L}^n(x) = \int_U u \, d\mathcal{L}^n(x) - a \int_U v \, d\mathcal{L}^n(x) = 0. \end{align*} [/step]

[step:Use positivity to force the zero-integral eigenfunction to vanish] Suppose, toward a contradiction, that $w \neq 0$. Since $w \in E_{\lambda_1}$ is a nonzero first eigenfunction, the positivity hypothesis in the theorem statement implies that either $w \geq 0$ $\mathcal{L}^n$-a.e. in $U$ with $w > 0$ on a set of positive $\mathcal{L}^n$-measure, or $w \leq 0$ $\mathcal{L}^n$-a.e. in $U$ with $w < 0$ on a set of positive $\mathcal{L}^n$-measure. Since $w \in E_{\lambda_1} \subset L^2(U)$ and the hypotheses on $U$ include boundedness, the [Cauchy-Schwarz Inequality](/theorems/432) applied to $|w| \cdot 1$ on $(U,\mathcal{B}(U),\mathcal{L}^n)$ gives $w \in L^1(U)$. A nonnegative $L^1$ function that is positive on a set of positive measure has strictly positive integral, because its integral is the supremum of the integrals of nonnegative simple functions dominated by it; applying this to $w$ and to $-w$ handles the two sign alternatives. In the first case, \begin{align*} \int_U w \, d\mathcal{L}^n(x) > 0, \end{align*} and in the second case, \begin{align*} \int_U w \, d\mathcal{L}^n(x) < 0. \end{align*} Both alternatives contradict \begin{align*} \int_U w \, d\mathcal{L}^n(x) = 0. \end{align*} Therefore $w = 0$ in $H_0^1(U)$, and hence $u = a v$. [/step]

[step:Conclude that the first eigenspace is one-dimensional] We have shown that any two nonzero elements of $E_{\lambda_1}$ are scalar multiples of one another. Since $\lambda_1$ is an eigenvalue, $E_{\lambda_1}$ contains at least one nonzero eigenfunction. Therefore $E_{\lambda_1}$ is a nonzero real vector space in which every two nonzero vectors are linearly dependent, so \begin{align*} \dim_{\mathbb R} E_{\lambda_1} = 1. \end{align*} This proves that the first Dirichlet eigenvalue is simple. [/step]