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.

Remarksi) (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 ($\mathcal{B}$

bwaand $\mathcal{B}^0$ha) known ? (combinatorial, easy to check examples are preferred)

.

Q2) Are there general statements ? ($\mathcal{B}$

bwa+some condition $\Longrightarrow$ $\mathcal{B}^0$ha)