Is there any book that proves the following theorem? The proof that is mentioned is only for covariant tensors and the second part is not done at all.
I think having Prop. B1 the proof of the seond part is rather short, isn't it? :
"$\implies$":
If $F$ is a $C^{\infty}$-mult.-linear map, then by Prop.B.1, $F$ can be identified with a (k+1,l)-tensor, namely, for some $\omega_{k+1}$,
$\omega_{k+1}((F\omega_{1},...,\omega_{k},v_{1},...v_{l}))$
"$\Longleftarrow$":
Given a (k+1,l)-tensor $\sigma$ Prop.B.1 says that $\Phi^{-1}[\sigma]$ is a $C^{\infty}$-mult.-linear map.
Proof attempt for Proposotion B.1:
First let's look hat the special case mentioned in excercsie B.2.
Let $\Phi:End(V)\to T^{(1,1)}(V),A\mapsto\Phi[A](\omega,v):=\omega(Av)$.
Claim: $\Phi$ is bijective
- Injective:
Suppose $\Phi[A]=\Phi[B]\iff\omega(Av)=\omega(Bv)\iff\omega(A-B)v)=0$, by linearity of $\omega$. Since this is true for all $\omega\in V^{*}$ and $v\in V$, it is also true for $dx_{i},e_{i}$, i.e. basis vectors of $V^{*}$ and $V$ respectively.
$(A-B)e_{j}$ is a vector with entries $a_{ij}-b_{ij}$.Thus,
$0=dx_{i}((A-B)e_{j})=a_{ij}-b_{ij},\forall i,j\iff A=B$.
- Surjective:
Since we deal with finite dimension surjectivity follows from the fact that $\Phi$ is injective.$dim V=dim Im(f)+dim\ker(f)$
"The longer way":
Let $\sigma\in T^{(1,1)}(V)$. We need to find $A$ such that $\Phi[A]=\sigma$.
Let's look at $\Phi[A](\omega,v)=\omega(Av) =\omega(\Sigma_{i=1}^{n}(\Sigma_{j=1}^{n}a_{ij}v_{j})e_{i}) =\Sigma_{ij}a_{ij}v_{j}\omega(e_{i}) =\Sigma_{ij}a_{ij}v_{j}\Sigma_{k}\omega^{k}dx_{k}(e_{i}) =\Sigma_{ij}a_{ij}v_{j}\omega^{i}$
Now let's look at $\sigma(\omega,v)=\sigma(\Sigma\omega^{i}dx_{i},\Sigma v^{j}e_{j})=\Sigma_{ij}\omega^{i}v^{j}\sigma(dx_{i},e_{j})$.
We want $\Phi[A](\omega,v)=\sigma(\omega,v)\iff\Sigma_{ij}a_{ij}v_{j}\omega^{i}=\Sigma_{ij}\omega^{i}v^{j}\sigma(dx_{i},e_{j})\iff\Sigma_{ij}(a_{ij}-\sigma(dx_{i},ej))v^{i}\omega^{j}=0$.
Since both $\omega,v\neq 0$ there have to be non-zero coefficients. Hence, $a_{ij}=\sigma(dx_{i},e_{j})$.
Therefore by the above calculations let $A:=(\sigma(dx_{i},e_{j}))$. Then the pretty much same calculation as in the very first one shows that $\Phi[A](\omega,v)=\sigma(\omega,v)$.
General case:
Since we still deal with finite dimensions it suffices to show injectivity:
Let $\Phi:Mult(V^{*}\times...\times V^{*}\times V\times...\times V,V)\to T^{(k+1,l)},\eta\mapsto\omega_{k+1}(\eta(\omega_{1},...,\omega_{k},v_{1},...v_{l}))$
Suppose $\Phi[\eta]=\Phi[\tau]\iff\omega_{k+1}(\eta)=\omega(\tau)\iff\omega_{k+1}(\eta-\tau)=0$ (the arugments are omitted).
Now as in the special case consider $\omega=dx_{i}$. Since $\eta-\tau\in V$ $dx_{i}(\eta-tau)=(\eta-\tau)_{i}$,i.e. the i-th component.
Thus, $0=dx_{i}(\eta-\tau)=(\eta-\tau)_{i}\iff \eta=\tau$. Hence, $\Phi$ is injective.
The reason why Ted is so confident in comments that the proof is exactly the same is because we have the following more general lemma, which generalizes both the covariant and contravariant cases. This is one of those times where generalizing makes everything simpler.
Below, $\Gamma(E)$ is the $C^\infty(B)$-module of sections of $E$.
Lemma. Let $E \to B$ and $F \to B$ be finite rank smooth real vector bundles (over a paracompact base). Every $C^\infty(B)$-linear map $\Gamma(E) \to \Gamma(F)$ is induced by a vector bundle map $E \to F$, and conversely.
Proof. Of course, a map of vector bundles $E \to F$ induces a $C^\infty(B)$-linear map of sections $\Gamma(E) \to \Gamma(F)$ by post-composition. Now suppose that $\phi : \Gamma(E) \to \Gamma(F)$ is $C^\infty(B)$-linear. We want to define a map $T : E \to F$ for each $e \in E_b$ by letting $\sigma \in \Gamma(E)$ be some section such that $\sigma(b) = e$ and then declaring $T(e) := \phi(\sigma)(b)$; it suffices to show that $T(e)$ defined this way is independent of the choice of section $\sigma$. In turn by $\mathbb{R}$-linearity it suffices to show that $\phi(\sigma)(b) = 0$ whenever $\sigma(b) = 0$.
We first claim that $\phi$ must be local, in the sense that if $\sigma \in \Gamma(E)$ is such that $\sigma = 0$ on some open neighbourhood $U \subset B$ of some $b \in B$, then $\phi(\sigma)(b) = 0$. But this is easy to see: pick a $C^\infty(B)$ function $f$ such that $f(b) = 1$ and $f = 0$ outside $U$. Then we must have $f \sigma = 0$ and so (by $\mathbb{R}$-linearity) $$ \phi(\sigma)(b) = f(b) \phi(\sigma)(b) = \phi(f \sigma)(b) = 0, $$ as claimed. Again by $\mathbb{R}$-linearity, this implies that if two sections $\sigma$ and $\sigma'$ agree on a neighbourhood of $b \in B$ then $\phi(\sigma)(b) = \phi(\sigma')(b)$.
Now fix any $\sigma \in \Gamma(E)$ and $b \in B$ for which $\sigma(b) = 0$, and pick an open neighbourhood $U$ on which $E$ and $F$ are each locally trivial. On $U$ there then exists a basis of sections $\{\alpha_i\}$ of $E$. This means that (on $U$) $$ \sigma(x) = \sum_i g_i(x) \alpha_i(x) $$ for some functions $g_i \in C^\infty(U)$. Note that since $\sigma(b) = 0$ this means that $g_i(b) = 0$ for all $i$. Choose some extension of the $a_i$s to all of $B$, and likewise extend the sections $\{\alpha_i\}$ to all of $B$ (in both cases this is possible by paracompactness of $B$, though the sections $\{\alpha_i\}$ need not be linearly independent outside of $U$). Then $\sigma$ and $\sum_i g_i \alpha_i$ agree on $U$ and therefore (by locality, then $C^\infty(B)$-linearity, then since $g_i(b) = 0$) $$ \phi(\sigma)(b) = \phi\left(\sum_i g_i \alpha_i\right)(b) = \sum_i g_i(b)\phi(\alpha_i)(b) = 0. $$ This completes the proof.
The lemma recovers your question by letting $E$ be the tensor product of $k$ copies of the tangent bundle of $M$ with $l$ copies of the cotangent bundle of $M$, and letting $F$ be the rank 1 trivial bundle $\mathcal{E}_1 = M \times \mathbb{R}$ over $M$ (since then $\Gamma(\mathcal{E}_1) = C^\infty(M)$).