 Solution

Write $X=\theta +Z$, with $Z\sim N(0,1)$ independent of the prior $\theta \sim N(0,{\tau }^{-1})$. Then $\lambda X-\theta =(\lambda -1)\theta +\lambda Z$, soCompleting the square in the joint density gives $\theta \mid X\sim N(X/(1+\tau ),1/(1+\tau ))$. Thus the Bayes rule is $d_{1/(1+\tau )}$, with Bayes risk $1/(1+\tau )$. The rule $d_1=X$ has integrated risk one for every $\tau >0$, so its excess Bayes risk is $\tau /(1+\tau )\to 0$ as $\tau \downarrow 0$. The priors remain proper at every positive $\tau$. Therefore $d_1$ is extended Bayes, without treating the limiting improper flat prior as a proper Bayes prior.