Tensor product of a line bundle with a large multiple of another positive line bundle also positive? Let $X$ be a complex manifold and $\mathcal{L}$ be a positive line bundle on $X$. If $E$ is any other line bundle on $X$, then is it true that for all sufficiently large $m$, $\mathcal{L}^m \otimes E$ is also positive?
When $X$ is compact, the answer is positive, and it follows by a standard compactness argument if you start with the definition that $\mathcal{L}$ is positive iff the Chern class $\omega$ of $\mathcal{L}$ satisfies: $\omega(x; v, Iv) > 0$ for all $x \in X$ and $v \in T_{\mathbb{R}, x}(X)$ (the real tangent space of $X$ at $x$) and $I: T_{\mathbb{R}, x}(X) \to T_{\mathbb{R}, x}(X)$ is the map induced by multiplication by $i$. 
So my real question is: is the above question true when $X$ is not compact? What if $X$ is an affine algebraic variety?
 A: Let us prove that for an affine variety $X$ every line bundle $E$ is "positive" according to the chosen defintion. All we need to prove is that for any hermitian metric $g$ on $E$ with curvature $w$ there is a Kahler form $w_1$ on $X$ such that $w_1>-w$. Since $X$ is affine, for any $w_1$ we have $w_1=\frac{i}{2\pi}\partial\bar\partial (f_1)$ and changing the metric $g$ on $E$ by $ge^{f_1}$ we corresponing curvature will change from $w$ to $w+w_1$, which we assume to be positive.
So we need to show the existence of arbitrary large $w_1$. Since $X$ is affine and hence admits an embedding in $\mathbb C^n$, it is enough to show this for $\mathbb C^n$. Moreover, since $\mathbb C^n=\mathbb C^1\times ...\times \mathbb C^1$ it is enought to prove the statement for $\mathbb C^1$. Now, on $\mathbb C^1$ every form of the shape $w_1=h_1dz\wedge d\bar z$ is Kahler for $h_1>0$ and we can chose $h_1$ as large as we wish. 
The conclusion is that if one choses this definition, then each line bundle on an affine $X$ is positive, which sounds strange. So I am not sure what should be a reasonable definition of positivincess in non-compact case, if it exists at all.
