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Yes, that doesn't make much of a differerence. For the finite-dimensional case the theorem is proved by induction (just in each step, take one element of a complementary subspace and expand the functional on it), and for the infinite dimensional as well, with the additional help of Zorn's lemma.

\Edit: Maybe its clearest if one realizes that multilinear functionals of n variables are nothing but linear functionals on the n-times tensor product of V. That way you can just apply the usual Hahn-Banach theorem.

\2nd edit: After thinking it through, I realized that it is actually not clear how to identify multisublinear functionals and sublinear functionals on the tensor product. So one is better off taking the other route.

In the proof of the usual Hahn-Banach theorem, one has to prove first that one can extend the given functional $T$ in the case of codimension one, i.e. if $\mathrm{dim} V/U = 1$, and then one proceeds by transfinite induction. If this is the case, every vector in $v \in V$ can be written in the form $$v = u + \lambda v_0$$ with $\lambda \in \mathbb{R}$ and $v_0 \in U$. Then one defines the extension $\tilde{T}$ by setting $$ \tilde{T}(v) = T(u) + \lambda r$$ where the $r$ is chosen in such a way, that $\tilde{T}$ is dominated by $S$. The main part of the proof is then the a bit lengthy calculation that such an $r$ exists. Now, if $T$ is a function of $n$ variables, this calculation gets even lenghtier and quite confusing.

Instead of fixing the original proof that way, it seems therefore easier to use the original theorem to prove that one can extend $T$ in the $k$th variable and then use induction on $k$.

To make this explicit, first choose an (Hamel) basis of $U$ and extend it to a basis of $V$ (denote this basis by $(b_i)_{i \in I}$). Let $T: U \times \dots \times U \longrightarrow \mathbb{R}$ be linear and $S: V \times \dots \times V \longrightarrow \mathbb{R}$ be sublinear such that $T \leq S$ on $U \times \dots \times U$.

Now it is possible to extend $T$ to a functional $\tilde{T}:V \times U \times \dots \times U \longrightarrow \mathbb{R}$ by using the original Hahn-Banach theorem on the linear functional $$ T( \cdot, b_{i_2}, \dots, b_{i_n}).$$ for each $n-1$-tuple $i_1, \dots, i_n$ in $I$ with the additional property that $b_{i_k} \in U$. Because the $(b_i)$ are linearly independant, this patches together to a linear map $\tilde{T}:V \times U \times \dots \times U \longrightarrow \mathbb{R}$.

Now suppose we have a linear map $T: V \times \dots \times V \times U \times \dots \times U \longrightarrow \mathbb{R}$ (her we have $k$ times $V$ and $n-k$ times $U$). Then one can use Hahn-Banach on the linear functionals $$ T(b_{i_1}, \dots, b_{i_k}, \cdot, b_{i_{k+2}}, \dots b_{i_n})$$ with any $i_1, \dots i_k \in I$ and $b_{i_{k+2}}, \dots b_{i_n} \in U$. Again, this patches together.

By induction, one has the extension on whole $V \times \dots \times V$.

In this answer I made more or less the same mistake over and over again. It turns out that sublinear functions are not as nice as I believed. I thought it would be best to delete it.

Yes, that doesn't make much of a differerence. For the finite-dimensional case the theorem is proved by induction (just in each step, take one element of a complementary subspace and expand the functional on it), and for the infinite dimensional as well, with the additional help of Zorn's lemma.

\Edit: Maybe its clearest if one realizes that multilinear functionals of n variables are nothing but linear functionals on the n-times tensor product of V. That way you can just apply the usual Hahn-Banach theorem.

Yes, that doesn't make much of a differerence. For the finite-dimensional case the theorem is proved by induction (just in each step, take one element of a complementary subspace and expand the functional on it), and for the infinite dimensional as well, with the additional help of Zorn's lemma.

\Edit: Maybe its clearest if one realizes that multilinear functionals of n variables are nothing but linear functionals on the n-times tensor product of V. That way you can just apply the usual Hahn-Banach theorem.

\2nd edit: After thinking it through, I realized that it is actually not clear how to identify multisublinear functionals and sublinear functionals on the tensor product. So one is better off taking the other route.

In the proof of the usual Hahn-Banach theorem, one has to prove first that one can extend the given functional $T$ in the case of codimension one, i.e. if $\mathrm{dim} V/U = 1$, and then one proceeds by transfinite induction. If this is the case, every vector in $v \in V$ can be written in the form $$v = u + \lambda v_0$$ with $\lambda \in \mathbb{R}$ and $v_0 \in U$. Then one defines the extension $\tilde{T}$ by setting $$ \tilde{T}(v) = T(u) + \lambda r$$ where the $r$ is chosen in such a way, that $\tilde{T}$ is dominated by $S$. The main part of the proof is then the a bit lengthy calculation that such an $r$ exists. Now, if $T$ is a function of $n$ variables, this calculation gets even lenghtier and quite confusing.

Instead of fixing the original proof that way, it seems therefore easier to use the original theorem to prove that one can extend $T$ in the $k$th variable and then use induction on $k$.

To make this explicit, first choose an (Hamel) basis of $U$ and extend it to a basis of $V$ (denote this basis by $(b_i)_{i \in I}$). Let $T: U \times \dots \times U \longrightarrow \mathbb{R}$ be linear and $S: V \times \dots \times V \longrightarrow \mathbb{R}$ be sublinear such that $T \leq S$ on $U \times \dots \times U$.

Now it is possible to extend $T$ to a functional $\tilde{T}:V \times U \times \dots \times U \longrightarrow \mathbb{R}$ by using the original Hahn-Banach theorem on the linear functional $$ T( \cdot, b_{i_2}, \dots, b_{i_n}).$$ for each $n-1$-tuple $i_1, \dots, i_n$ in $I$ with the additional property that $b_{i_k} \in U$. Because the $(b_i)$ are linearly independant, this patches together to a linear map $\tilde{T}:V \times U \times \dots \times U \longrightarrow \mathbb{R}$.

Now suppose we have a linear map $T: V \times \dots \times V \times U \times \dots \times U \longrightarrow \mathbb{R}$ (her we have $k$ times $V$ and $n-k$ times $U$). Then one can use Hahn-Banach on the linear functionals $$ T(b_{i_1}, \dots, b_{i_k}, \cdot, b_{i_{k+2}}, \dots b_{i_n})$$ with any $i_1, \dots i_k \in I$ and $b_{i_{k+2}}, \dots b_{i_n} \in U$. Again, this patches together.

By induction, one has the extension on whole $V \times \dots \times V$.

Yes, that doesn't make much of a differerence. For the finite-dimensional case the theorem is proved by induction (just in each step, take one element of a complementary subspace and expand the functional on it), and for the infinite dimensional as well, with the additional help of Zorn's lemma.

Yes, that doesn't make much of a differerence. For the finite-dimensional case the theorem is proved by induction (just in each step, take one element of a complementary subspace and expand the functional on it), and for the infinite dimensional as well, with the additional help of Zorn's lemma.

\Edit: Maybe its clearest if one realizes that multilinear functionals of n variables are nothing but linear functionals on the n-times tensor product of V. That way you can just apply the usual Hahn-Banach theorem.