Is every maximal ideal in a C*-algebra always closed? I wonder if every maximal two-sided (self-adjoint) ideal in a C*-algebra is automatically closed. It is a very basic fact of C*-algebra theory that it holds true for the unital case. In the non-unital case, there certainly exists a non-closed dense ideal but the point is that such an ideal may never be maximal. It is a non-trivial fact that the answer is affirmative for the commutative case [D. Rudd, On isomorphisms between ideals in rings of continuous functions. Trans. Amer. Math. Soc. 159 (1971) 335--353].  
One can ask a similar question for maximal *-subalgebras. 
 A: Theorem: Let $J\subseteq A$ be a maximal ideal. Then $J$ is hereditary (if $a\in A_+$ satisfies $a\leq b$ for some $b\in J_+$, then $a\in J_+$), strongly invariant (if $x^*x\in J_+$ then $xx^*\in J_+$) and if $a\in J_+$ then $a^t\in J_+$ for every $t>0$.
We need some notation and some lemmas first:
We define the relation $\sim$ on $A_+$ by setting $a\sim b$ if there exists $x\in A$ such that $a=xx^*$ and $b=x^*x$.
Lemma 1: If $a\leq b\sim b'$, then there exists $a'$ such that $a\sim a'\leq b'$.
Proof: Let $x\in A$ such that $b=x^*x$ and $b'=xx^*$.
Let $x=v|x|$ be the polar decomposition in $A^{**}$.
Set $y:=va^{1/2}$ and $a':=yy^*$.
Then $y$ belongs to $A$ and we have $a=y^*y$ and $a'=yy^*=vav^*\leq vbv^*=b'$.
Lemma 2: Let $a,b,c\in A_+$ and $t>1$ such that $a\sim b\leq c^t$.
Then there exists $y\in A$ such that $a=ycy^*$.
Proof: Choose $x\in A$ such that $a=xx^*$ and $b=x^*x$.
Set $\alpha:=\tfrac{1}{2t}$.
Then $0<\alpha<1/2$.
Applying the polar decomposition in C*-algebras (see Proposition~II.3.2.1 in Blackadar's Operator algebras book) we obtain $y\in A$ such that $x = y(c^t)^\alpha$.
Then
$$ a = xx^* = yc^{2t\alpha}y^* = ycy^*. $$
Proof of Theorem:
We consider the ideal as suggested by @Black
$$ K := \{ a\in A : a^*a \leq b \text{ for some } b\in J_+ \}. $$
We will show that $J=K$.
Indeed, as noted by @Black, we either have $J=K$ or $K=A$.
We show that $K=A$ leads to a contradiction.
So assume that $K=A$.
Let $a\in A_+$.
Since $a^{1/4}\in K$, we obtain $b\in J_+$ such that $a^{1/2}\leq b$.
Then
$$ a = a^{1/4}a^{1/2}a^{1/4} \leq a^{1/4}ba^{1/4} \sim b^{1/2}a^{1/2}b^{1/2} \leq b^2. $$
By Lemma 1, we obtain $a'$ such that $a\sim a'\leq b^2$.
It then follows from Lemma 2 that $a\in J_+$.
Thus $J=A$, a contradiction.
With an argument as in the answer of @Black, it now follows that $J$ is hereditary, strongly invariant, and closed under roots of positive elements.
A: Since this question has been around for over three years without an answer, I suppose it is OK to give a partial answer even if the advance is tiny.
THEOREM: Let $J$ be a maximal ideal in a C*-algebra $A$.  Then either
(i) $J$ is hereditary and for every $a\in J_+$, and every $r>0$, one has that $a^r\in J$, or
(ii) $A_+$ coincides with its order-ideal generated by $J_+$.
PROOF: It is easy to see that $M:= \{m\in A:\exists b\in J,\ m^*m\leq b\}$ is an ideal containing $J$.  Given that $J$ is maximal we conclude that $M=J$ or $M=A$.  In the first case we claim that $J$ is hereditary.  To see this suppose that $0\leq a\leq b$, where $a\in A$ and $b\in J$.  Setting $m=a^{1/2}$, we then have that $$ m^*m = a \leq b \in J, $$ so $m$ is in $M$, hence also in $J$, so $a=m^2\in J$, proving that $J$ is hereditary.
Still under the assumption that $M=J$, let us prove the last part of (i).  For this let $a\in J_+$ and put $m=a^{1/2}$.  Then $$ m^*m=a\leq a\in J, $$ so we see that $m\in M=J$.  In other words $a^{1/2}\in J$.  By iterating this procedure we conclude that $a^{1/2^n}\in J$, for all $n$.  Given $r>0$, pick a sufficiently large integer $n$ such that $1/2^n<r$.  Then $$ a^r=a^{1/2^n}a^{r-1/2^n}\in J.$$
In the second case, namely if $M=A$, we will prove that the order-ideal generated by $J_+$ coincides with $A_+$.  To see this, let $a$ be an arbitrary positive element in $A$.  Then $m:= a^{1/2}$ lies in $M$ by hypothesis, so there is some $b$ such that $$ a= m^*m\leq b\in J, $$ proving that $a$ lies in the order-ideal generated by $J_+$, as desired.

REMARK.  Not all ideals in a C*-algebra are hereditary.  See Is every 2-sided ideal in a C*-algebra hereditary?
