5 subscripted function correctly

Let $\langle \operatorname{Ent},+,\cdot \rangle$ be the (complex) vector space of entire functions.
For all members $n$ of $\{1,2,3,...\}$, define $||\cdot ||_n : \operatorname{Ent} \to \mathbb{R}$ by $||f||_n = \operatorname{sup}(\{|f(z)| : |z|\leq n\})$.
$\big\langle \operatorname{Ent},+,\cdot,\{||.||_n : n\in \{1,2,3,...\}\} \big\rangle$ is a Frechet space.

For all complex numbers $z_0$ and members $g$ of $\operatorname{Ent}$, the operators $L_1,...,L_5 L_1,...,L_4 : \operatorname{Ent} \to \operatorname{Ent}$ defined by

$(i) \quad (L_1(f))(z) = g(z)\cdot f(z)$

$(ii) \quad (L_2(f))(z) = f(z+z_0)$ f(g(z))(iii) \quad (L_3(f))(z) = f(0)$f'(z)$

$(iv) \quad (L_4(f))(z) = f'(z)$

$(v) \quad (L_5(f))(z) = \displaystyle\int_0^z f$

are all continuous and linear.

Let $S$ be the set of all functions obtainable by the above.
Let $\mathbf{L}$ be continuous operator algebra on $\operatorname{Ent}$.
Let $T$ be the closure of $S$ as a sub-algebra of $\mathbf{L}$.

Does $\:$ $T = \mathbf{L}$ $\:$ ?
If no, is $T$ dense in $\mathbf{L}$? (uniform operator topology)
If no again, is $T$ dense in $\mathbf{L}$ in some weaker topology?

4 Rollback to Revision 2

Let $\langle \operatorname{Ent},+,\cdot \rangle$ be the (complex) vector space of entire functions.
For all members $n$ of $\{1,2,3,...\}$, define $||\cdot ||_n : \operatorname{Ent} \to \mathbb{R}$ by $||f||_n = \operatorname{sup}(\{|f(z)| : |z|\leq n\})$.
$\big\langle \operatorname{Ent},+,\cdot,\{||.||_n : n\in \{1,2,3,...\}\} \big\rangle$ is a Frechet space.

For all complex numbers $z_0$ and members $g$ of $\operatorname{Ent}$, the operators $L_1,...,L_6 L_1,...,L_5 : \operatorname{Ent} \to \operatorname{Ent}$ defined by

$(i) \quad (L_1(f))(z) = g(z)\cdot f(z)$

$(ii) \quad (L_2(f))(z) = f(z+z_0)$

$(iii) \quad (L_3(f))(z) = f(0)$

$(iv) \quad (L_4(f))(z) = f(g(z))$ f'(z)(v) \quad (L_5(f))(z) = f'(z)(vi) \quad (L_6(f))(z) = \displaystyle\int_0^z f$are all continuous and linear. Let$S$be the set of all functions obtainable by the above. Let$\mathbf{L}$be continuous operator algebra on$\operatorname{Ent}$. Let$T$be the closure of$S$as a sub-algebra of$\mathbf{L}$. Does$\:T = \mathbf{L}\:$? If no, is$T$dense in$\mathbf{L}$? (uniform operator topology) If no again, is$T$dense in$\mathbf{L}$in some weaker topology? 3 added composition Let$\langle \operatorname{Ent},+,\cdot \rangle$be the (complex) vector space of entire functions. For all members$n$of $\{1,2,3,...\}$, define$||\cdot ||_n : \operatorname{Ent} \to \mathbb{R}$by $||f||_n = \operatorname{sup}(\{|f(z)| : |z|\leq n\})$. $\big\langle \operatorname{Ent},+,\cdot,\{||.||_n : n\in \{1,2,3,...\}\} \big\rangle$ is a Frechet space. For all complex numbers$z_0$and members$g$of$\operatorname{Ent}$, the operators$L_1,...,L_5 L_1,...,L_6 : \operatorname{Ent} \to \operatorname{Ent}$defined by$(i) \quad (L_1(f))(z) = g(z)\cdot f(z)(ii) \quad (L_2(f))(z) = f(z+z_0)(iii) \quad (L_3(f))(z) = f(0)(iv) \quad (L_4(f))(z) = f'(z)$f(g(z))$

$(v) \quad (L_5(f))(z) = f'(z)$

$(vi) \quad (L_6(f))(z) = \displaystyle\int_0^z f$

are all continuous and linear.

Let $S$ be the set of all functions obtainable by the above.
Let $\mathbf{L}$ be continuous operator algebra on $\operatorname{Ent}$.
Let $T$ be the closure of $S$ as a sub-algebra of $\mathbf{L}$.

Does $\:$ $T = \mathbf{L}$ $\:$ ?
If no, is $T$ dense in $\mathbf{L}$? (uniform operator topology)
If no again, is $T$ dense in $\mathbf{L}$ in some weaker topology?

2 changed "constant" to "complex numbers"
1