It is known from the general theory that, given a bialgebra (over a field $k$)
\begin{equation}
\mathcal{B}=(B,\mu,1\_B,\Delta,\epsilon)
\end{equation} 
the Sweedler's dual $\mathcal{B}^0$ (called also Hopf or restricted dual, i.e. the space of linear functionals $f\in B^\*$ such that $f\circ\mu\in \mathcal{B}^\*\otimes\_k \mathcal{B}^\*$) is a bialgebra (with the transposed elements). It seems that it can happen that $\mathcal{B}$ admit no antipode whereas $\mathcal{B}^0$ does (and then be a Hopf algebra). My evidence (however neither an example nor a counterexample) is the following : 



> 
> Let $k$ be a field and $q\in k^\times$. On the usual algebra of polynomials, $(k[x],.,1)$, let us define a structure of bialgebra by 
>  $$
> \Delta(x)=x\otimes 1+ 1\otimes x + q x\otimes x 
> $$ 
>  one checks easily that $\mathcal{B}=(k[x],.,1,\Delta,\epsilon)$ ($\epsilon(P)=P(0)$ as usual) is a bialgebra and, as $g=(1+qx)$ (which is group-like) has no inverse, $\mathcal{B}$ is a bialgebra without antipode. However all the elements of $\mathcal{B}$ are dualizable (indeed, the dual of $\Delta$ is the infiltration product of Chen, Fox and Lyndon, Free differential calculus) within $k[x]$ and the bialgebra so obtained $\mathcal{B}^\vee=(k[x],\mu\_\Delta,1,\Delta\_{conc},\epsilon)$ admits an antipode. 
> 
> 
> **Remarks** i) (Foissy). The Hopf algebra $\mathcal{B}^\vee$ can be considered as a subbialgebra of $\mathcal{B}^0$ (the full Sweedler's dual) which is NOT a Hopf algebra. To see this, remark that the characters of $\mathcal{B}$ are in $\mathcal{B}^0$ and the values of a character $\chi$ are fixed by $\chi(x)$. Call $F\_a$ be the character s.t. $F\_a(x)=a$. Convolution of characters satisfies $F\_a\*F\_b(x)=a+b+qab$. Now, for all $b\in k$, we have $F\_{-1/q}\*F\_b=F\_{-1/q}$ which proves that the set of characters (i.e. group-like elements of $\mathcal{B}^0$) is NOT a group under convolution. 
> 
> 
> ii) The comultiplication above was introduced, for $q=1$, by Chen, Fox, and Lyndon, Free differential calculus IV, Ann. Math. 1958. It can be defined on a noncommutative alphabet $X$ (i.e. to complete as a bialgebra a free algebra $A\langle X\rangle$) by 
>  $$
> \Delta\_{\uparrow}(x):=x\otimes 1+ 1\otimes x + q x\otimes x
> $$
>  forall $x\in X$. remark that it is compatible with all types of commutation between the letters. 
> 
> 
> The comultiplication of such an infiltration bialgebra $(A\langle X\rangle,conc,1\_{X^\*},\Delta\_{\uparrow},\epsilon)$ has a pretty combinatorial expression. It reads, on a word $w\in X^\*$,
>  $$
> \Delta\_{\uparrow}(w)=\sum\_{I\cup J=[1..n]}\,q^{|I\cap J|}w[I]\otimes w[J]
> $$
>  where $n=|w|$ is the length of the word.
> 
> 
> 


My questions are the following (*bwa*=bialgebra without antipode, *ha*=Hopf algebra)



> 
> Q1) Are there significant families of examples of the situation ($\mathcal{B}$ *bwa* and $\mathcal{B}^0$ *ha*) known ? (combinatorial, easy to check examples are preferred)
> 
> 
> 


.



> 
> Q2) Are there general statements ? ($\mathcal{B}$ *bwa* +some condition $\Longrightarrow$ $\mathcal{B}^0$ *ha*) 
> 
> 
> 



Bialgebras with Hopf restricted (or Sweedler) duals
Actually your guess is true. Consider the *left* Hopf algebra $\widetilde{SL\_q(2)}$ from [Iyer, Taft, *The dual of a certain left quantum group*](https://doi.org/10.1142/S0219498816500729). This is constructed essentially as $SL\_q(2)$, but imposing just half of the relations. Namely, if $SL\_q(2)$ is given by $\Bbbk\langle X\_{i,j}\mid i,j=1,2\rangle$ with relations
\begin{gather}
X\_{21}X\_{11}=qX\_{11}X\_{21}, \tag{$\star$}\\
X\_{12}X\_{11}=qX\_{11}X\_{12}, \\
X\_{22}X\_{12}=qX\_{12}X\_{22}, \tag{$\*$}\\
X\_{22}X\_{21}=qX\_{21}X\_{22}, \\
X\_{22}X\_{11}=qX\_{12}X\_{21}+1, \tag{$\circ$} \\
X\_{21}X\_{12}=qX\_{11}X\_{22}-q\cdot 1, \tag{$\bullet$} \\
X\_{21}X\_{12}=X\_{12}X\_{21}, \\
\end{gather}
then $\widetilde{SL\_q(2)}$ is obtained by just requiring $(\star),(\*),(\circ)$ and $(\bullet)$. It turns out that $\widetilde{SL\_q(2)}$ admits a left convolution inverse of the identity which is not a right convolution inverse (hence it is not an antipode). However, its Sweedler dual coincides with $SL\_q(2)^\circ$, whence it is Hopf.


