the lexicographic order becomes $x_1\prec \cdots \prec x_d\prec x_{d+1}$), and construct for each $m_i$ $(i=1,\cdots,n)$ a monomial $\tilde{m}_i\in\mathcal{M}[{\pmb{x}},x_{d+1}]$ such that $\tilde{m}_i=m_ix_{d+1}^{c_i}$, where $c_i=f(i)-\deg(m_i)\in{\mathbb{N}}$. So $\deg(\tilde{m}_i)=f(i)$ and thus the restriction (\[equ:special\]) is satisfied by $\tilde{m}_1,\cdots,\tilde{m}_n$. Furthermore,

$\tilde{m}_1,\cdots,\tilde{m}_n$ is a $f$-generating sequence of ${\mathbb{K}}[{\pmb{x}},x_{d+1}]$.

*Proof:* It suffices to prove that $\tilde{m}_{i+1}\notin\langle\tilde{m}_1,\cdots,\tilde{m}_{i}\rangle$ for every $i=1,2,\cdots,n-1$. In fact, if it is not the case then $\tilde{m}_{i+1}$ should be a multiple of some $\tilde{m}_j$, where $j\leq i$. It implies that $m_{i+1}$ is a multiple of $m_j$, which is contrary to $m_{i+1}\notin\{m_1,\cdots,m_i\}$. $\Box$

Then it immediately follows from the definition of $\Omega$ that $n\leq\Omega(d,f,f(1))$. Since $m_1,\cdots,m_n$ is arbitrarily chosen, we obtain $L(d,f)\leq\Omega(d,f,f(1))$ from Proposition \[prop:monomial\]. Conversely, we also show that $\Omega(d,f,f(1))\leq L(d,f)$. We consider the sequence $\hat{m}_1,\cdots,\hat{m}_N$ of ${\mathbb{K}}[{\pmb{x}},x_{d+1}]$, which is defined as in equation (\[equ:hatm\]). Then $N=\Omega(d,f,f(1))$. By putting $x_{d+1}=1$, this sequence becomes another sequence $m_1^\prime,\cdots,m_N^\prime$ of ${\mathbb{K}}[{\pmb{x}}]$.

$m_1^\prime,\cdots,m_n^\prime$ is a $f$-generating sequence.

*Proof:* Observe that $\deg(m_i^\prime)\leq\deg(\tilde{m}_i)=f(i)$ for all $i$. Then we only need to prove that $m_i^\prime\notin\{m_1^\prime,\cdots,m_{i-1}^\prime\}$ for all $i$. We assume that $m_i^\prime$ is a multiple of some $m_j^\prime$, where $j<i$. Let $\tilde{m}_i=m_i^\prime x_{d+1}^\alpha$ and $\tilde{m}_j=m_j^\prime x_{d+1}^\beta$, then $\alpha<\beta$ (otherwise $\tilde{m}_i$ would be a multiple of $\tilde{m}_j$). We also have $f(j)<f(i)$; otherwise,f(j)=f(i) and 