I would like to expand the polynomial $p(\lambda) = \sum_{i=0}^{d} a_{i} \lambda^{i}$, as $F(p(\lambda), \lambda_{0})= \min_{j} [ val(a_{j}) + j \lambda_{0} ] $ with $\lambda_{0}$ being a real variable. Here, $val(a)$ is the valuation, which is the lowest exponent of the non-zero Puiseux series. The valuation satisfies \begin{align} val(a) =\infty \text{ iff } a=0 ,\quad val(ab) =val(a)+val(b),\quad val(a+b) \geq \min [ val(a) , val(b) ]. \end{align} For example, $val(t^{2} -2t+3) = \min [ val(t^2) , val(-2t), val(3) ] =\min [ 2,1,0 ]=0$. The above suggested expansion for polynomial $p=\lambda^{2} - x^{2} -2x$ with respect to variable $\lambda$ reads $F(p(\lambda), \lambda_{0}) = \min[1, 2 \lambda_{0}]$.

For a simple polynomial, I have obtained the valuation using

poly = -2 x  + x^2 + 3
val = Min[Exponent[DeleteCases[MonomialList[poly], {0}], x]]

However, this fails when MonomialList has zero elements. I have excluded these cases using DeleteCases such that

poly = -2 x  - x^2 +  \[Lambda]^2
coeff = CoefficientList[poly, \[Lambda]]
val = Min @@@ Exponent[DeleteCases[MonomialList[coeff], {0}], x]
(*{1, 0}*)

The output val has components for $\lambda^{0}$ and $\lambda^{2}$ and $\lambda^{1}$ is the deleted case. Now, the remaining challenge is how to properly implement $F(p, \lambda_{0})$. Do you have any suggestions on obtaining this function?

I would like to expand the polynomial $p(\lambda) = \sum_{i=0}^{d} a_{i} \lambda^{i}$, as $F(p(\lambda), \lambda_{0})= \min_{j} [ val(a_{j}) + j \lambda_{0} ] $ with $\lambda_{0}$ being a real variable. Here, $val(a)$ is the valuation, which is the lowest exponent of the non-zero Puiseux series. The valuation satisfies \begin{align} val(a) =\infty \text{ iff } a=0 ,\quad val(ab) =val(a)+val(b),\quad val(a+b) \geq \min [ val(a) , val(b) ]. \end{align} For example, $val(t^{2} -2t+3) = \min [ val(t^2) , val(-2t), val(3) ] =\min [ 2,1,0 ]=0$. The above suggested expansion for polynomial $p=\lambda^{2} - x^{2} -2x$ with respect to variable $\lambda$ reads $F(p(\lambda), \lambda_{0}) = \min[1, 2 \lambda_{0}]$.

For a simple polynomial, I have obtained the valuation using

poly = -2 x  + x^2 + 3
val = Min[Exponent[DeleteCases[MonomialList[poly], {0}], x]]

However, this fails when MonomialList has zero elements. I have excluded these cases using DeleteCases such that

poly = -2 x  - x^2 +  \[Lambda]^2
coeff = CoefficientList[poly, \[Lambda]]
val = Min @@@ Exponent[DeleteCases[MonomialList[coeff], {0}], x]
(*{1, 0}*)

The output val has components for $\lambda^{0}$ and $\lambda^{2}$ and $\lambda^{1}$ is the deleted case. To obtain $F$ for the particular example, I have used

F[i_, val_] := Min[val + i Subscript[\[Lambda], 0]];
F[{0, 2}, val ]
(*{1, 2 Subscript[\[Lambda], 0]}*)

Here, I have now explicitly passed to the function that $\ i \in \{0,2\}$ as $i=0$ was excluded. 

Do you have any suggestions on better implementing these steps so that $F$ can be calculated more easily?