butions with means and covariance matrices characterized by (27-30). To complete the proof of Theorem 1  we need to establish the tightness of ln L ( h ; λ ) and ln L ( h ; µ )  viewed as random elements of the space C [ 0   h ] $^{r}$  as n  p → ∞ so that p/n → c. Formulae (25-26) and the facts that S − p = O$_{p}$ (1)   and that Δ$_{p}$ ( z$_{i}$$_{0}$ ) = O$_{p}$ (1) for i = 1   ...  r  where O$_{p}$ (1) are uniform in h ∈ ( 0   ¯ h ] $^{r}$  imply that for an arbitrarily small positive ε  there must exist B > 0 such that Pr ( sup$_{h}$$_{∈}$ ( 0   ¯ h ] r | ln L ( h ; λ ) | > B ) < ε and Pr ( sup$_{h}$$_{∈}$ ( 0   ¯ h ] r | ln L ( h ; µ ) | > B ) < ε for sufficiently large n and p. Since  as implied by Proposition 1  ln L ( h ; λ ) and ln L ( h ; µ ) are continuous functions on h ∈ [ 0   h ] $^{r}$  sup$_{h}$$_{∈}$ ( 0   ¯ h ] r | ln L ( h ; λ ) | = sup$_{h}$$_{∈}$ [ 0   ¯ h ] r | ln L ( h ; λ ) |   and sup$_{h}$$_{∈}$ ( 0   ¯ h ] r | ln L ( h ; µ ) | = sup$_{h}$$_{∈}$ [ 0   ¯ h ] r | ln L ( h ; µ ) |   so that the tightness of ln L ( h ; λ ) and ln L ( h ; µ ) follows. □ To save space  we only derive the asymptotic power envelope for the relatively more difficult case of real-valued data and µ -based tests. According to the Neyman- Pearson lemma  the most powerful test of the null h = 0 against a point alternative h = ( h$_{1}$  ...  h$_{r}$ ) is the test which rejects the null when ln L ( h ; µ ) is larger than a critical value C. It follows from Theorem 1 that  for such a test to have asymptotic size α   C must be where Now  according to Le Cam’s third lemma and Theorem 1  under h = ( h$_{1}$  ...  h$_{r}$ )   ln L ( h ; µ ) d → N ( m ( h ) + W ( h )   W ( h )) . Therefore  the asymptotic power β$_{µ}$ ( h ) m ( h ) = 1 2 r ∑ i j =1 ( ln ( 1 − h$_{i}$h$_{j}$ c ) + h$_{i}$h$_{j}$ c ) and W ( h ) = − r ∑ i j =1 ( ln ( 1 − h$_{i}$h$_{j}$ c ) + h$_{i}$h$_{j}$ c ) . C = √ W ( h )Φ − $^{1}$(1 − α ) + m ( h )   (63) Proof of theorem 2 37