= Solution
Let $F_0(T)$ be the observed initial <instantaneous forward rate> curve. Since $\widehat W_0=0$, part (i) gives $f_0(T)=g(T)-\sigma^2T^2/2$. Thus the <forward-curve calibration of a shifted Brownian short rate> is achieved by the deterministic choice
$$
\boxed{g(T)=F_0(T)+\frac{\sigma^2T^2}{2}.}
$$
Substitution gives $f_0(T)=F_0(T)$ at every maturity where the curve is defined classically. It also reproduces the corresponding initial <zero-coupon bond> curve:
$$
P_0(T)=\exp\left[-\int_0^TF_0(s)ds-\frac{\sigma^2T^3}{6}+\frac{\sigma^2T^3}{6}\right]=\exp\left[-\int_0^TF_0(s)ds\right].
$$
There is no shape restriction on the initial curve beyond the regularity needed to define these integrals and forward <derivatives>; for continuous curves the match is pointwise, and for locally integrable curves the forward identity has its almost-everywhere interpretation.
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