[proofplan] We realize the Gaussian copula through a bivariate normal random vector and rewrite the upper tail coefficient as a quotient of joint and marginal Gaussian tail probabilities. A linear orthogonal-normal transformation diagonalizes the covariance structure and reduces the joint upper tail event to a one-dimensional standard normal tail with a larger threshold. Comparing this exponential upper bound with a matching elementary lower bound for the marginal normal tail proves that the quotient tends to zero. The lower tail follows by applying the same upper-tail estimate to $(-Z_1,-Z_2)$, which has the same correlation. [/proofplan]

[step:Represent the copula tails as Gaussian tail probabilities]Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space supporting a centered Gaussian random vector \begin{align*} Z: \Omega &\to \mathbb{R}^2\\ \omega &\mapsto (Z_1(\omega),Z_2(\omega)) \end{align*} with covariance matrix \begin{align*} \Sigma_r = \begin{pmatrix} 1 & r\\ r & 1 \end{pmatrix}. \end{align*} For $u \in (0,1)$, define $x_u := \Phi^{-1}(u) \in \mathbb{R}$. Since each coordinate has standard normal distribution, \begin{align*} 1-u = \mathbb{P}(Z_1 > x_u). \end{align*} Also, \begin{align*} 1 - 2u + C_r(u,u) &= \mathbb{P}(Z_1 > x_u, Z_2 > x_u). \end{align*} Indeed, $C_r(u,u)=\mathbb{P}(Z_1 \le x_u,Z_2 \le x_u)$, and inclusion-exclusion gives \begin{align*} \mathbb{P}(Z_1 > x_u, Z_2 > x_u) &= 1-\mathbb{P}(Z_1 \le x_u)-\mathbb{P}(Z_2 \le x_u)+\mathbb{P}(Z_1 \le x_u,Z_2 \le x_u)\\ &= 1-2u+C_r(u,u). \end{align*} Thus \begin{align*} \frac{1 - 2u + C_r(u,u)}{1-u} = \frac{\mathbb{P}(Z_1 > x_u,Z_2 > x_u)}{\mathbb{P}(Z_1 > x_u)}. \end{align*} As $u \uparrow 1$, $x_u \to \infty$, so it remains to prove \begin{align*} \lim_{x \to \infty} \frac{\mathbb{P}(Z_1 > x,Z_2 > x)}{\mathbb{P}(Z_1 > x)} =0. \end{align*}[/step]

[guided]The copula is defined by applying the bivariate Gaussian distribution to normal quantiles. Therefore the diagonal value $C_r(u,u)$ is exactly the probability that both Gaussian coordinates lie below the same threshold $x_u=\Phi^{-1}(u)$: \begin{align*} C_r(u,u)=\mathbb{P}(Z_1 \le x_u,Z_2 \le x_u). \end{align*} The upper tail coefficient uses the expression $1-2u+C_r(u,u)$. This expression is not accidental: by inclusion-exclusion it is the probability of the simultaneous upper tail event. Since $\mathbb{P}(Z_1 \le x_u)=\mathbb{P}(Z_2 \le x_u)=u$, we have \begin{align*} \mathbb{P}(Z_1 > x_u,Z_2 > x_u) &= 1-\mathbb{P}(Z_1 \le x_u)-\mathbb{P}(Z_2 \le x_u)+\mathbb{P}(Z_1 \le x_u,Z_2 \le x_u)\\ &= 1-2u+C_r(u,u). \end{align*} Similarly, because $Z_1$ is standard normal, \begin{align*} 1-u=\mathbb{P}(Z_1>x_u). \end{align*} Thus the upper tail coefficient is the limit of a conditional-type quotient: \begin{align*} \frac{1 - 2u + C_r(u,u)}{1-u} = \frac{\mathbb{P}(Z_1 > x_u,Z_2 > x_u)}{\mathbb{P}(Z_1 > x_u)}. \end{align*} The limit $u \uparrow 1$ corresponds exactly to $x_u=\Phi^{-1}(u)\to\infty$. Hence the problem becomes a comparison between a bivariate Gaussian joint tail and a one-dimensional Gaussian tail.[/guided]

[step:Diagonalize the correlated Gaussian vector]Let $U: \Omega \to \mathbb{R}$ and $V: \Omega \to \mathbb{R}$ be the real-valued random variables \begin{align*} U &= \frac{Z_1+Z_2}{\sqrt{2(1+r)}},& V &= \frac{Z_1-Z_2}{\sqrt{2(1-r)}}. \end{align*} Since $(Z_1,Z_2)$ is centered Gaussian, every linear image is Gaussian. Direct covariance computation gives \begin{align*} \mathbb{E}[U] &= 0,& \mathbb{E}[V] &= 0,\\ \operatorname{Var}(U) &= \frac{\operatorname{Var}(Z_1)+2\operatorname{Cov}(Z_1,Z_2)+\operatorname{Var}(Z_2)}{2(1+r)} =1,\\ \operatorname{Var}(V) &= \frac{\operatorname{Var}(Z_1)-2\operatorname{Cov}(Z_1,Z_2)+\operatorname{Var}(Z_2)}{2(1-r)} =1,\\ \operatorname{Cov}(U,V) &= \frac{\operatorname{Var}(Z_1)-\operatorname{Var}(Z_2)}{2\sqrt{(1+r)(1-r)}}=0. \end{align*} Because a centered Gaussian vector with diagonal covariance has independent coordinates, $U$ and $V$ are independent standard normal random variables. Solving for $Z_1$ and $Z_2$ gives \begin{align*} Z_1 &= \frac{\sqrt{1+r}\,U+\sqrt{1-r}\,V}{\sqrt{2}},\\ Z_2 &= \frac{\sqrt{1+r}\,U-\sqrt{1-r}\,V}{\sqrt{2}}. \end{align*} Therefore, for $x>0$, \begin{align*} \{Z_1>x,\ Z_2>x\} \subset \left\{U>\sqrt{\frac{2}{1+r}}\,x\right\}. \end{align*} Indeed, adding the two inequalities $Z_1>x$ and $Z_2>x$ gives $\sqrt{2(1+r)}\,U>2x$.[/step]

[guided]The dependence between $Z_1$ and $Z_2$ is carried by their sum and difference. Define \begin{align*} U &= \frac{Z_1+Z_2}{\sqrt{2(1+r)}},& V &= \frac{Z_1-Z_2}{\sqrt{2(1-r)}}. \end{align*} The denominators are chosen so that both variables have variance $1$. We verify this explicitly: \begin{align*} \operatorname{Var}(U) &= \frac{\operatorname{Var}(Z_1)+2\operatorname{Cov}(Z_1,Z_2)+\operatorname{Var}(Z_2)}{2(1+r)} = \frac{1+2r+1}{2(1+r)} =1,\\ \operatorname{Var}(V) &= \frac{\operatorname{Var}(Z_1)-2\operatorname{Cov}(Z_1,Z_2)+\operatorname{Var}(Z_2)}{2(1-r)} = \frac{1-2r+1}{2(1-r)} =1. \end{align*} Their covariance is \begin{align*} \operatorname{Cov}(U,V) &= \frac{\operatorname{Cov}(Z_1+Z_2,Z_1-Z_2)}{2\sqrt{(1+r)(1-r)}}\\ &= \frac{\operatorname{Var}(Z_1)-\operatorname{Cov}(Z_1,Z_2)+\operatorname{Cov}(Z_2,Z_1)-\operatorname{Var}(Z_2)}{2\sqrt{(1+r)(1-r)}}\\ &=0. \end{align*} Since $(U,V)$ is a centered Gaussian vector, zero covariance implies independence. Thus $U$ and $V$ are independent standard normal variables. Now invert the linear transformation: \begin{align*} Z_1 &= \frac{\sqrt{1+r}\,U+\sqrt{1-r}\,V}{\sqrt{2}},\\ Z_2 &= \frac{\sqrt{1+r}\,U-\sqrt{1-r}\,V}{\sqrt{2}}. \end{align*} If both $Z_1>x$ and $Z_2>x$, then adding the two inequalities eliminates $V$: \begin{align*} Z_1+Z_2 > 2x. \end{align*} But $Z_1+Z_2=\sqrt{2(1+r)}\,U$, so \begin{align*} U>\sqrt{\frac{2}{1+r}}\,x. \end{align*} This gives the event inclusion \begin{align*} \{Z_1>x,\ Z_2>x\} \subset \left\{U>\sqrt{\frac{2}{1+r}}\,x\right\}. \end{align*} The important point is that $\sqrt{2/(1+r)}>1$ whenever $r<1$, so the joint event forces a strictly more extreme one-dimensional Gaussian tail.[/guided]

[step:Compare the joint upper tail with the marginal upper tail]Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define the standard normal density $\varphi: \mathbb{R} \to (0,\infty)$ by \begin{align*} \varphi(t)=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}. \end{align*} For $a:=\sqrt{2/(1+r)}>1$ and $x>0$, the previous step gives \begin{align*} \mathbb{P}(Z_1>x,Z_2>x) \le \mathbb{P}(U>ax). \end{align*} Using the elementary upper Gaussian tail bound \begin{align*} \mathbb{P}(U>y) = \int_y^\infty \varphi(t)\,d\mathcal{L}^1(t) \le \frac{1}{y\sqrt{2\pi}}e^{-y^2/2}, \qquad y>0, \end{align*} with $y=ax$, we obtain \begin{align*} \mathbb{P}(Z_1>x,Z_2>x) \le \frac{1}{ax\sqrt{2\pi}}e^{-a^2x^2/2}. \end{align*} For the marginal lower bound, since $\varphi$ is decreasing on $[0,\infty)$, \begin{align*} \mathbb{P}(Z_1>x) = \int_x^\infty \varphi(t)\,d\mathcal{L}^1(t) \ge \int_x^{x+1/x}\varphi(t)\,d\mathcal{L}^1(t) \ge \frac{1}{x}\varphi\left(x+\frac{1}{x}\right) \end{align*} for $x>0$. Since \begin{align*} \left(x+\frac{1}{x}\right)^2 = x^2+2+\frac{1}{x^2} \le x^2+3 \end{align*} for $x \ge 1$, we have \begin{align*} \mathbb{P}(Z_1>x) \ge \frac{e^{-3/2}}{x\sqrt{2\pi}}e^{-x^2/2}, \qquad x\ge 1. \end{align*} Combining the two estimates yields, for $x\ge1$, \begin{align*} 0 \le \frac{\mathbb{P}(Z_1>x,Z_2>x)}{\mathbb{P}(Z_1>x)} \le \frac{e^{3/2}}{a} \exp\left(-\frac{a^2-1}{2}x^2\right). \end{align*} Because \begin{align*} a^2-1 = \frac{2}{1+r}-1 = \frac{1-r}{1+r} >0, \end{align*} the right-hand side tends to $0$ as $x\to\infty$. Hence \begin{align*} \lim_{x\to\infty} \frac{\mathbb{P}(Z_1>x,Z_2>x)}{\mathbb{P}(Z_1>x)} =0. \end{align*}[/step]

[guided]Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. The event inclusion from the previous step reduces the joint tail to a one-dimensional Gaussian tail: \begin{align*} \mathbb{P}(Z_1>x,Z_2>x) \le \mathbb{P}\left(U>\sqrt{\frac{2}{1+r}}\,x\right). \end{align*} Set \begin{align*} a:=\sqrt{\frac{2}{1+r}}. \end{align*} Since $r<1$, we have $a>1$. The standard Gaussian tail estimate gives, for $y>0$, \begin{align*} \mathbb{P}(U>y) = \int_y^\infty \frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,d\mathcal{L}^1(t) \le \frac{1}{y\sqrt{2\pi}}e^{-y^2/2}. \end{align*} This estimate follows by writing, for $t\ge y>0$, $1\le t/y$, and therefore \begin{align*} \int_y^\infty e^{-t^2/2}\,d\mathcal{L}^1(t) \le \frac{1}{y}\int_y^\infty t e^{-t^2/2}\,d\mathcal{L}^1(t) = \frac{1}{y}e^{-y^2/2}. \end{align*} Applying this with $y=ax$ gives \begin{align*} \mathbb{P}(Z_1>x,Z_2>x) \le \frac{1}{ax\sqrt{2\pi}}e^{-a^2x^2/2}. \end{align*} We also need a lower bound for the denominator $\mathbb{P}(Z_1>x)$, because the quotient could only be controlled if the marginal tail is not underestimated. Since the standard normal density \begin{align*} \varphi: \mathbb{R} &\to (0,\infty)\\ t &\mapsto \frac{1}{\sqrt{2\pi}}e^{-t^2/2} \end{align*} is decreasing on $[0,\infty)$, for $x>0$ we restrict the integration domain from $(x,\infty)$ to the smaller interval $(x,x+1/x)$ and obtain \begin{align*} \mathbb{P}(Z_1>x) = \int_x^\infty \varphi(t)\,d\mathcal{L}^1(t) \ge \int_x^{x+1/x}\varphi(t)\,d\mathcal{L}^1(t) \ge \frac{1}{x}\varphi\left(x+\frac{1}{x}\right). \end{align*} For $x\ge1$, \begin{align*} \left(x+\frac{1}{x}\right)^2 = x^2+2+\frac{1}{x^2} \le x^2+3, \end{align*} so \begin{align*} \mathbb{P}(Z_1>x) \ge \frac{e^{-3/2}}{x\sqrt{2\pi}}e^{-x^2/2}. \end{align*} Dividing the joint upper bound by this marginal lower bound gives \begin{align*} 0 \le \frac{\mathbb{P}(Z_1>x,Z_2>x)}{\mathbb{P}(Z_1>x)} \le \frac{e^{3/2}}{a} \exp\left(-\frac{a^2-1}{2}x^2\right). \end{align*} The exponent is genuinely negative because \begin{align*} a^2-1 = \frac{2}{1+r}-1 = \frac{1-r}{1+r} >0. \end{align*} This strict positivity is exactly where the assumption $r<1$ is used. Therefore the right-hand side tends to $0$, and the desired upper-tail quotient tends to $0$.[/guided]

[step:Conclude that the upper tail dependence coefficient is zero] Since $x_u=\Phi^{-1}(u)\to\infty$ as $u\uparrow1$, the limit from the previous step gives \begin{align*} \lambda_U &= \lim_{u\uparrow1} \frac{1-2u+C_r(u,u)}{1-u}\\ &= \lim_{u\uparrow1} \frac{\mathbb{P}(Z_1>x_u,Z_2>x_u)}{\mathbb{P}(Z_1>x_u)} =0. \end{align*} Thus the Gaussian copula has no upper tail dependence for every $r\in(-1,1)$. [/step]

[step:Apply the same upper-tail estimate to the reflected vector for the lower tail] For $u\in(0,1)$, again write $x_u=\Phi^{-1}(u)$. Then \begin{align*} C_r(u,u) = \mathbb{P}(Z_1\le x_u,Z_2\le x_u). \end{align*} Define the reflected Gaussian vector $W:\Omega\to\mathbb{R}^2$ by \begin{align*} W(\omega)=(-Z_1(\omega),-Z_2(\omega)). \end{align*} The vector $W=(W_1,W_2)$ is centered Gaussian with unit variances and correlation $r$. If $u\downarrow0$, then $x_u\to-\infty$ and $y_u:=-x_u\to\infty$. Therefore \begin{align*} \frac{C_r(u,u)}{u} &= \frac{\mathbb{P}(Z_1\le x_u,Z_2\le x_u)}{\mathbb{P}(Z_1\le x_u)}\\ &= \frac{\mathbb{P}(W_1\ge y_u,W_2\ge y_u)}{\mathbb{P}(W_1\ge y_u)}. \end{align*} Replacing non-strict inequalities by strict inequalities does not change these probabilities because one-dimensional and two-dimensional Gaussian distributions assign probability zero to affine hyperplanes. Applying the already proved upper-tail estimate to $W$ yields \begin{align*} \lim_{u\downarrow0}\frac{C_r(u,u)}{u}=0. \end{align*} Thus $\lambda_L=0$. Combining this with $\lambda_U=0$ proves \begin{align*} \lambda_L=\lambda_U=0. \end{align*} [/step]