The efficient frontier in mean variance criterion

Question

The efficient frontier is the portfolios with the minimum of variance ($V$) at a given mean ($E$) or a maximum of mean at a given variance，Why do the optimal portfolios in the effcient frontier, is the efficient frontier consistent with the maximum of expected utility?
For example, assume a logarithmic utility function U=log(1+R), through Taylor series expansion and keep the first three terms,  we can get the expected utility $EU=E-(E^2+V)/2$, from this equation, take the partial derivative of $E$, it is not monotonical increase, it is the same with $V$, it seems the efficient frontier does not lead to the maximum of expected utility.
I understand if the utility function is expoential, the expected utility $EU=E-lambda V$, which is exactly consistent with the efficient frontiers. I just do not know the other form of utility function, like the logarithmic form as above.
Any help would be appreciated.

Alecos Papadopoulos · Answer

Using $mathbb E$ for the expected value symbol, $E_R$ and $V_R$ for the mean and the avriance of returns $R$, for a utility function of the form
$$U(R) = ln(1+R)$$
the second-order Taylor expansion gives (ignoring the remainder)
$$U(R) approx ln(1+E_R) + frac{1}{1+E_R} (R- E_R) -frac{1}{2(1+E_R)^2}(R-E_R)^2.$$
Then
$$mathbb E[U(R)] approx E_R + 0 - frac{V_R}{2(1+E_R)^2}. $$
We then have
$$frac{partial mathbb E[U(R)] }{partial V_R} = -frac{1}{2(1+E_R)^2} < 0$$
so decreasing the variance always increases expected utility, while
$$frac{partial mathbb E[U(R)] }{partial E_R} = 1 + frac{V_R}{(1+E_R)^3} <0$$.
The reason why this derivative is always positive, is that even with negative returns, by construction "returns" cannot fall below $-100% = -1$.
So increasing mean returns always increases expected utility.
I don't see where the problem is. What am I missing?

The efficient frontier in mean variance criterion

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