To show there exists a unique function $u \in C^{1}(\mathbb{C^n})$ that satisfies $(\bar{\partial u})=f$ Assume $n \gt 1$. Let $f$ be a $(0,1)$ form in $\mathbb{C^n}$, with $C^1$-coefficients and compact support $K$, such that $\bar{\partial} f=0$. Let $\Omega_{0}$ be the unbounded component of $\mathbb{C^n}-K$. I need to show there exists a unique function $u \in C^{1}(\mathbb{C^n})$ that satisfies $(\bar{\partial u})=f$ as well as $u(z)=0$ for every $z \in \Omega_{0}$
Let $f=\sum f_j(z) d\bar{z_j} $. Let's define for $z \in \mathbb{C^n}$ $$u(z)=\frac{1}{2\pi i}\int_{\mathbb{C}}f_{1}(\lambda,z_2,..,z_n) \frac{d\lambda \wedge d\bar{\lambda}}{\lambda-z_1}$$
$$=\frac{1}{2\pi i}\int_{\mathbb{C}}f_{1}(\lambda+z_1,z_2,..,z_n) \frac{d\lambda \wedge d\bar{\lambda}}{\lambda}$$
I am trying to evoke the following proposition here: 

Let $\Omega \subset \mathbb{C}$ be a bounded open set. Suppose $f \in C^{1
}(\Omega)$, $f$ is bounded and $$u(z)=\frac{1}{2\pi i} \int_{\Omega} \frac{f(\lambda)}{\lambda-z} d\lambda \wedge d\bar{\lambda}, (z \in \Omega).$$ Then $u \in C^{1}(\Omega)$ and $\bar{D}u=f$.

From the above mentioned Proposition, $\bar{D_1}u=f_1$. I also think that $u \in C^1(\mathbb{C})$ follows from here but I am not sure about it. Now since we have $(\bar{\partial f})=0$, for $2 \le j \le n$, $\bar{D_j}f_1=\bar{D_1}f_j$, so that $$(\bar{D_j})u(z)=\frac{1}{2\pi i}\int_{\mathbb{C}}(\bar{D_j}f_{1})(\lambda,z_2,..,z_n) \frac{d\lambda \wedge d\bar{\lambda}}{\lambda-z_1}$$(why can I take the differential inside: because the function is bounded ??)
$$=(\bar{D_j})u(z)=\frac{1}{2\pi i}\int_{\mathbb{C}}(\bar{D_1}f_{j})(\lambda,z_2,..,z_n) \frac{d\lambda \wedge d\bar{\lambda}}{\lambda-z_1}$$
It should follow from here that the last statement is equal to $f_j(z)$. I am thinking of this proposition (but not able to find out why)

$$\text{Let $\Omega$ be a bounded region in $\mathbb{C}$, with smooth oriented boundary $\partial\Omega$. If $u \in C^1(\bar{\Omega})$, then we have $u(a)=\dfrac{1}{2\pi i}\int_{\partial \Omega}\dfrac{u(\lambda)d\lambda}{\lambda-a}-\dfrac{1}{2\pi i}\int_{\Omega}\dfrac{\bar{D}u(\lambda)d\bar{\lambda}\wedge d\lambda}{\lambda-a} $}$$

Suppose $\bar{D_j}u=f_j$ for $1 \le j \le n$, then we are done with the first part. This also shows that $u$ is holomorphic in $\Omega_0$(as the derivatives vanish there). I have no idea how to show that $u(z)=0$ in $\Omega_0$( take $|z_1|$ large in the first equation??)
Also why this theorem fails when $n=1$??
Thanks for the help
 A: 
From the above mentioned Proposition, $\overline{D}_1u=f_1$. I also think that $u \in C^1(\mathbb{C})$ follows from here but I am not sure about it.

Yes, the assertion follows from the proposition, but we need to add a few arguments, since the situation we have here is slightly different. In the proposition, we have a bounded $\Omega \subset \mathbb{C}$ while here we look at the entire space. But since $f_1$ has compact support, we can restrict the integral to a bounded $\Omega$ without changing $u$, as long as $\Omega$ contains the support of $f_1$. Additionally, in the proposition the integrand is a function of a single variable, while here $f_1$ depends on several variables. The theorems about parameter-dependent integrals yield the desired conclusion. However, in my opinion it is nicest to use the alternative form
$$u(z) = \frac{1}{2\pi i} \int_{\mathbb{C}} f_1(\lambda + z_1,z_2,\dotsc,z_n)\cdot \frac{d\lambda\wedge d\overline{\lambda}}{\lambda}$$
of the integral to deduce that $u \in C^1(\mathbb{C}^n)$. Since $f_1 \in C^1(\mathbb{C}^n)$ and $f_1$ has compact support, the theorems about differentiation under the integral - whether for Riemann integrals or Lebesgue integrals - immediately show that $u \in C^1(\mathbb{C}^n)$ and
$$\frac{\partial u}{\partial w}(z) = \frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\partial f_1}{\partial w}(\lambda + z_1,z_2,\dotsc,z_n)\cdot \frac{d\lambda\wedge d\overline{\lambda}}{\lambda},$$
where $\frac{\partial}{\partial w}$ can be any of the real partial derivatives $\frac{\partial}{\partial x_k},\,\frac{\partial}{\partial y_k}$ or the Wirtinger derivatives $\frac{\partial}{\partial z_k},\, \frac{\partial}{\partial \overline{z}_k}$.
We then invoke the proposition only to see that - for fixed $z_2,\dotsc,z_n$ - we have $\frac{\partial u}{\partial \overline{z}_1} = f_1$.

I am thinking of this proposition (but not able to find out why)

We use that proposition (the inhomogeneous Cauchy integral formula/Cauchy-Pompeiu formula) for the function $g\colon z_1 \mapsto f_j(z_1,z_2,\dotsc,z_n)$ with $z_2,\dotsc,z_n$ fixed, and $\Omega$ a large disk containing the support of $g$. [Since $f_j$ has compact support, its slice $g$ also has compact support, for arbitrary fixed $z_2,\dotsc,z_n$. Instead of a disk, we could take any other domain with smooth boundary containing the support of $g$, naturally.] By the inhomogeneous Cauchy integral formula, we have
\begin{align}
f_j(z_1,z_2,\dotsc,z_n) &= g(z_1)\\
&= \frac{1}{2\pi i} \int_{\partial \Omega} \frac{g(\lambda)}{\lambda - z_1}\,d\lambda - \frac{1}{2\pi i} \int_{\Omega} \frac{\partial g}{\partial \overline{z}_1}(\lambda)\frac{d\overline{\lambda}\wedge d\lambda}{\lambda - z_1}\\
&= - \frac{1}{2\pi i} \int_{\Omega} \frac{\partial g}{\partial \overline{z}_1}(\lambda)\frac{d\overline{\lambda}\wedge d\lambda}{\lambda - z_1} \tag{$\operatorname{supp g} \subset \Omega$}\\
&= \frac{1}{2\pi i} \int_{\Omega} \frac{\partial g}{\partial \overline{z}_1}(\lambda)\frac{d\lambda\wedge d\overline{\lambda}}{\lambda - z_1}\\
&= \frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\partial g}{\partial \overline{z}_1}(\lambda)\frac{d\lambda\wedge d\overline{\lambda}}{\lambda - z_1} \tag{$\operatorname{supp g} \subset \Omega$}\\
&= \frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\partial f_j}{\partial \overline{z}_1} (\lambda,z_2,\dotsc,z_n)\frac{d\lambda\wedge d\overline{\lambda}}{\lambda - z_1}\\
&= \frac{1}{2\pi i} \int_{\mathbb{C}} \frac{\partial f_1}{\partial \overline{z}_j} (\lambda,z_2,\dotsc,z_n)\frac{d\lambda\wedge d\overline{\lambda}}{\lambda - z_1} \tag{$\frac{\partial f_i}{\partial \overline{z}_j} = \frac{\partial f_j}{\partial \overline{z}_i}$}\\
&= \frac{\partial u}{\partial \overline{z}_j}(z)
\end{align}
for any such $\Omega$.

I have no idea how to show that $u(z)=0$ in $\Omega_0$( take $\lvert z_1\rvert$ large in the first equation?)

Not quite. We integrate over the first variable, so that one is inconvenient to play with (and, as we see below, playing with $z_1$ wouldn't lead to the conclusion). The other variables are just parameters for the integral, so we can more easily play with them. Since $f_1$ has compact support, there is an $M \in (0,+\infty)$ such that $f_1(z) = 0$ for all $z$ with $\lvert z_2\rvert > M$. Hence we have $u(z) = 0$ for $\lvert z_2\rvert > M$, since the integrand in
$$u(z) = \frac{1}{2\pi i}\int_{\mathbb{C}} \frac{f_1(\lambda,z_2,\dotsc,z_n)}{\lambda - z_1}\,d\lambda \wedge d\overline{\lambda}$$
vanishes identically for such $z$. By definition of $\Omega_0$, the set $\{ z \in \Omega_0 : \lvert z_2\rvert > M\}$ is a nonempty open subset of $\Omega_0$, and $\Omega_0$ is connected, hence by the identity theorem the holomorphic function $u\lvert_{\Omega_0}$ is identically $0$.

Also why this theorem fails when $n=1$?

Only one part of the theorem fails for $n = 1$. We generally don't have $u(z) = 0$ for $z \in \Omega_0$ then. For $f \in C^1(\mathbb{C})$ with compact support, the function
$$u(z) = \frac{1}{2\pi i}\int_{\mathbb{C}} \frac{f(\lambda)}{\lambda - z}\,d\lambda\wedge d\overline{\lambda}$$
is a continuously differentiable function with $\frac{\partial u}{\partial \overline{z}} = f$ also for $n = 1$. It just generally doesn't have compact support. 
We could say the reason is because for $n = 1$ we have no parameter we can manipulate to make the integral vanish for an open subset of "large" $z$.
A more theory-oriented reason is that Hartogs' Kugelsatz doesn't hold for $n = 1$:
Suppose we have a holomorphic function $g$ defined on $G = \{ z \in \mathbb{C}^n : \lVert z\rVert > M\}$ for some $M \in (0,+\infty)$. Let $\varphi \in C^{\infty}_c(\mathbb{C}^n)$ be a cut-off function such that $\varphi(z) = 1$ for $\lVert z\rVert \leqslant M+1$ and $\operatorname{supp} \varphi \subset \{ z : \lVert z\rVert \leqslant M+2\}$. Then $(1-\varphi(z))\cdot g(z)$ is a smooth function on $\mathbb{C}^n$ if we define it to be $0$ on $\mathbb{C}^n\setminus G$, since $1-\varphi$ vanishes identically on a neighbourhood of $\mathbb{C}^n\setminus G$. Now define
$$f_j(z) = \frac{\partial}{\partial \overline{z}_j} \bigl((1-\varphi)\cdot g\bigr)(z).$$
Since $(1-\varphi)\cdot g$ is holomorphic for $\lVert z\rVert < M+1$ and $\lVert z\rVert > M+2$, $\omega := \sum_j f_j(z)\,d\overline{z}_j$ is a compactly supported smooth $\overline{\partial}$-closed $(0,1)$-form. If we have $\omega = \overline{\partial} u$ for a compactly supported $u$, then
$$h(z) = (1-\varphi(z))\cdot g(z) - u(z)$$
is an entire holomorphic function, and since $u$ has compact support, we have $h(z) = g(z)$ for all large enough $\lVert z\rVert$. By the identity theorem, since $G$ is connected, it follows that $h\lvert_G \equiv g$, so $h$ is an extension of $g$ to an entire function.
Since for $n = 1$ holomorphic functions can have non-removable isolated singularities (and worse), it follows that for $n = 1$ not all compactly supported smooth $(0,1)$-forms [$\overline{\partial}$-closed is automatic for $n = 1$] can be of the form $\overline{\partial} u$ for compactly supported $u$.
