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For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound above the $L^\infty$ norm of $f$ by a positive constant times the $L^1$ norm of the $n$-th order derivative of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality.

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound the $L^\infty$ norm of $f$ from above by a positive constant times the sum of the $L^1$ norms of the $n$-th order derivatives of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality.

(side remark: the case of Sobolev's embedding theorem which is relevant to the present context is an easy consequence of the fundamental theorem of Calculus applied once to each variable of $f$, which also shows that the constant above may be taken equal to 1)

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound above the $L^\infty$ norm of $f$ by a positive constant times the $L^1$ norm of the $n$-th order derivative of $f$, where $n$ is the dimension of the domain $\Omega$. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$.

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound above the $L^\infty$ norm of $f$ by a positive constant times the $L^1$ norm of the $n$-th order derivative of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality.

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. Assuming, for instance, that all Taylor coefficients at a given point are positive, one can also arrange the $L^1$ norms of the derivatives of the function $f$ provided by Borel's theorem to blow up arbitrarily fast. Since you assumed the domain $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$.

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound above the $L^\infty$ norm of $f$ by a positive constant times the $L^1$ norm of the $n$-th order derivative of $f$, where $n$ is the dimension of the domain $\Omega$. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$.