Let $K$ a field and $A$ subalgebra of $S=K[x_{1}, \ldots, x_{n}]$. If $<$ is a monomial order in $S$, we say that $in_{<}(A)$ is the $K$-subálgebra of $S$ generated over $K$ by all monomials $in_{<}(f)$ with $f \in A$. $in_{<}(A)$ is called initial algebra of $A$. $\mathcal{S} \subset A$ is called a Sagbi basis of $A$ if the elements $in(f)$ with $f \in \mathcal{S}$ generated the $K$-álgebra $in_{<}(A)$.
My question is, do you know an example of an algebra $A$ finitely generated but $in_{<}(A)$ not? Thanks for your answers!
My favorite example is the ring invariant of the alternating group $A_3$ with lex order. We can show that this subalgebra $A= \mathbb C[x,y,z]^{A_3}$ is finitely generated (in fact, it is $ \mathbb C[s_1,s_2,s_3,s]$ where $s_1=x+y+z,s_2=xy+yz+zx, s_3=xyz$ and $s=(x-y)(y-z)(z-x)$).
Now what is a basis (as a vector space) for $A$? We can take the following basis: $x^ay^bz^c+x^by^cz^a+x^cy^az^b$ where $a\ge b$ and $a\ge c$. With lex order, the leading term of this element is $x^ay^bz^c$. We can show that the algebra $in_{<}(A)$ is not finitely generated by showing that the elements of the form $x^{n+1}z^n$, $n=1,2,\ldots $ (which are leading term of some element in $A$ when $b=0$) can not be generated by those elements of the form $x^ay^bz^c$.
In fact, if $x^{n+1}z^n=(x^az^b)(x^cz^d)$ then since there is $f,g \in A$ such that $in_{<}(f)=x^az^b$ and $in_{<}(g)=x^cz^d$. Let $\sigma =(123) \in A_3$, we know that $f=\sigma(f)$ so $\sigma(x^az^b)=y^ax^b$ is a monomial in $f$. Since we are using lex order, we have $x^az^b>x^by^a$, hence, $a\ge b+1$. Similarly, $c\ge d+1$. Thus $$n+1=a+c\ge b+d+2=n+2$$ which is a contradiction.