Construction of a line bundle with lower degree and lower dimension of global sections Let $\mathcal{L}$ be a line bundle on a smooth projective irreducible curve $C/k=\overline{k}$ with genus $g$ and $\text{deg}\mathcal{L}=g$. 
Assume $\text{dim}H^0(C,\mathcal{L})>1$. 
It's my aim to construct a line bundle $\mathcal{L^{\prime}}$ on $C$ such that 
$\text{deg}\mathcal{L^{\prime}}=g-1$ and $\text{dim}H^0(C,\mathcal{L^{\prime}})=\text{dim}H^0(C,\mathcal{L})-1$.
My idea was to take a closed point $p$ in $C$ and define $\mathcal{L^{\prime}}:=\mathcal{L}\otimes \mathcal{O}(-[p])$. Then, we had $\text{deg}\mathcal{L^{\prime}}=g-1$ but I don't know how to argue for the dimension of the global sections. Does someone have an idea?
 A: The trick to this question is that we need to choose the point $p$ wisely. For the moment, I will choose $p$ to be a non-basepoint of your line bundle $\mathcal{L}$. Consider the short exact sequence of sheaves 
$$0 \to \mathcal{O}_C(-p) \to \mathcal{O}_C \to k(p) \to 0$$
where $k(p)$ is the skyscraper sheaf at $p$. Tensoring this with $\mathcal{L}$, we get a short exact sequence
$$0 \to \mathcal{L}\otimes \mathcal{O}_C(-p) \to \mathcal{L} \to k(p) \to 0.$$ Notice the right hand side is still $k(p)$ as $\mathcal{L}$ is locally free of rank $1$. Now consider the LES in cohomology
$$ 0 \to H^0(C,\mathcal{L}\otimes \mathcal{O}_C(-p)) \to H^0(C,\mathcal{L}) \to H^0(C,k(p))\to \\
 \hspace{50mm} H^1(C,\mathcal{L} \otimes \mathcal{O}_C(-p)) \to H^1(C,\mathcal{L}) \to 0.$$
The $0$ on the right appears because the higher cohomology of a skyscraper sheaf is zero. Computing dimensions, your result will follow once we show 
$$h^1(C,\mathcal{L}\otimes\mathcal{O}_C(-p)) = h^1(C,\mathcal{L}).$$
Apply Riemann-Roch  to the line bundles $\mathcal{L}$ and $\mathcal{L}\otimes \mathcal{O}_C(-p)$ to obtain
$$\begin{eqnarray} h^1(C,\mathcal{L}\otimes \mathcal{O}_C(-p)) &=&h^1(C,\mathcal{L}\otimes \mathcal{O}_C(-p)) \hspace{5mm} (1)\\
h^1(C,\mathcal{L}) &=& h^0(C,\mathcal{L}) - 1\hspace{17.5mm} (2). \end{eqnarray}$$
The left sides of (1) and (2) are equal if and only if $h^0(C,\mathcal{L}\otimes \mathcal{O}_C(-p)) = h^0(C,\mathcal{L}) - 1$, which holds since $p$ was chosen not to be a basepoint of $\mathcal{L}$.
