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Edit: According to the comment of Konstantinos, I add this link as a motivation for this question:

http://bims.iranjournals.ir/article_872_fd7287eb7f1365d9156e9da3ccb25196.pdf

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

Edit: According to the comment of Konstantinos, I add this link as a motivation for this question:

http://bims.iranjournals.ir/article_872_fd7287eb7f1365d9156e9da3ccb25196.pdf

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

Edit: According to the comment of Konstantinos, I add this link as a motivation for this question:

http://bims.iranjournals.ir/article_872_fd7287eb7f1365d9156e9da3ccb25196.pdf

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is that does the above finctional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$

What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?

Note that the differentiation operator on these coalgebras satisfies the above functional equation.

Our next question is the following: Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$