I think I have a proof.

To show the right side is contained in the left, let $x \in R$ be a homogeneous element for which there exists $n \in \mathbb{Z}$ such that $x^nM \subset N$. Given any $m \in M$ break it up into homogeneous pieces, $m=m_1+\ldots+m_s$ say. Each $x^nm_i$ lies in a different homogeneous piece of $M$ by the graded structure and since $x$ is homogeneous. But the annihilator condition implies it also lies in $N$ and $N^{\ast}$ by definition - it is the submodule generated by all homogeneous elements. Thus $x$ belongs to the left hand side.

For the opposite inclusion, suppose that for some $x \in R$ there exists $n \in \mathbb{Z}$ such that $x^nM \subset N^{\ast}$. Since $N \supset N^{\ast}$ it is sufficient to show that each homogeneous piece of $x$ lies in the right hand side. I think induction on the number of homogeneous pieces is a good way to go. The base case is clear.

Let $x=x_1+x_2\ldots+x_s$ be a homogeneous decomposition where $x_1$ is the highest degree piece, with $(x_2 + \ldots +x_s)$ belonging to the right hand side (that is, each $x_i$ with $i \geq 2$ lies in the right hand side). Let $m \in M$ be a homogeneous element. Then $x^nm=x_1^nm + l.o.t$ where the other terms have smaller degree. Since $x^nm \in N^{\ast}$ each homogeneous piece must be too. $x_1^nm$ lies in a higher degree piece than everything else and so this power of $x_1$ works, since every element $M$ is a sum of such homogeneous pieces.

EDIT: This is also posted at SE, see here. Apparently the definition I used for the radical is not correct - instead it is the intersection of all the prime ideals containing the object. I would imagine that the above is not longer useful.

I think I have a proof.

To show the right side is contained in the left, let $x \in R$ be a homogeneous element for which there exists $n \in \mathbb{Z}$ such that $x^nM \subset N$. Given any $m \in M$ break it up into homogeneous pieces, $m=m_1+\ldots+m_s$ say. Each $x^nm_i$ lies in a different homogeneous piece of $M$ by the graded structure and since $x$ is homogeneous. But the annihilator condition implies it also lies in $N$ and $N^{\ast}$ by definition - it is the submodule generated by all homogeneous elements. Thus $x$ belongs to the left hand side.

For the opposite inclusion, suppose that for some $x \in R$ there exists $n \in \mathbb{Z}$ such that $x^nM \subset N^{\ast}$. Since $N \supset N^{\ast}$ it is sufficient to show that each homogeneous piece of $x$ lies in the right hand side. I think induction on the number of homogeneous pieces is a good way to go. The base case is clear.

Let $x=x_1+x_2\ldots+x_s$ be a homogeneous decomposition where $x_1$ is the highest degree piece, with $(x_2 + \ldots +x_s)$ belonging to the right hand side (that is, each $x_i$ with $i \geq 2$ lies in the right hand side). Let $m \in M$ be a homogeneous element. Then $x^nm=x_1^nm + l.o.t$ where the other terms have smaller degree. Since $x^nm \in N^{\ast}$ each homogeneous piece must be too. $x_1^nm$ lies in a higher degree piece than everything else and so this power of $x_1$ works, since every element $M$ is a sum of such homogeneous pieces.

EDIT: This is also posted at SE, see here. Apparently the definition I used for the radical is not correct - instead it is the intersection of all the prime ideals containing the object. I would imagine that the above is not longer useful.

I think I have a proof.

To show the right side is contained in the left, let $x \in R$ be a homogeneous element for which there exists $n \in \mathbb{Z}$ such that $x^nM \subset N$. Given any $m \in M$ break it up into homogeneous pieces, $m=m_1+\ldots+m_s$ say. Each $x^nm_i$ lies in a different homogeneous piece of $M$ by the graded structure and since $x$ is homogeneous. But the annihilator condition implies it also lies in $N$ and $N^{\ast}$ by definition - it is the submodule generated by all homogeneous elements. Thus $x$ belongs to the left hand side.

For the opposite inclusion, suppose that for some $x \in R$ there exists $n \in \mathbb{Z}$ such that $x^nM \subset N^{\ast}$. Since $N \supset N^{\ast}$ it is sufficient to show that each homogeneous piece of $x$ lies in the right hand side. I think induction on the number of homogeneous pieces is a good way to go. The base case is clear.

Let $x=x_1+x_2\ldots+x_s$ be a homogeneous decomposition where $x_1$ is the highest degree piece, with $(x_2 + \ldots +x_s)$ belonging to the right hand side (that is, each $x_i$ with $i \geq 2$ lies in the right hand side). Let $m \in M$ be a homogeneous element. Then $x^nm=x_1^nm + l.o.t$ where the other terms have smaller degree. Since $x^nm \in N^{\ast}$ each homogeneous piece must be too. $x_1^nm$ lies in a higher degree piece than everything else and so this power of $x_1$ works, since every element $M$ is a sum of such homogeneous pieces.

I think I have a proof.

To show the right side is contained in the left, let $x \in R$ be a homogeneous element for which there exists $n \in \mathbb{Z}$ such that $x^nM \subset N$. Given any $m \in M$ break it up into homogeneous pieces, $m=m_1+\ldots+m_s$ say. Each $x^nm_i$ lies in a different homogeneous piece of $M$ by the graded structure and since $x$ is homogeneous. But the annihilator condition implies it also lies in $N$ and $N^{\ast}$ by definition - it is the submodule generated by all homogeneous elements. Thus $x$ belongs to the left hand side.

For the opposite inclusion, suppose that for some $x \in R$ there exists $n \in \mathbb{Z}$ such that $x^nM \subset N^{\ast}$. Since $N \supset N^{\ast}$ it is sufficient to show that each homogeneous piece of $x$ lies in the right hand side. I think induction on the number of homogeneous pieces is a good way to go. The base case is clear.

Let $x=x_1+x_2\ldots+x_s$ be a homogeneous decomposition where $x_1$ is the highest degree piece, with $(x_2 + \ldots +x_s)$ belonging to the right hand side (that is, each $x_i$ with $i \geq 2$ lies in the right hand side). Let $m \in M$ be a homogeneous element. Then $x^nm=x_1^nm + l.o.t$ where the other terms have smaller degree. Since $x^nm \in N^{\ast}$ each homogeneous piece must be too. $x_1^nm$ lies in a higher degree piece than everything else and so this power of $x_1$ works, since every element $M$ is a sum of such homogeneous pieces.

EDIT: This is also posted at SE, see here. Apparently the definition I used for the radical is not correct - instead it is the intersection of all the prime ideals containing the object. I would imagine that the above is not longer useful.