[proofplan] We first check that derivations form an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$. Then we prove that the commutator of two derivations is again a derivation by expanding both compositions and cancelling the mixed terms. The Lie algebra identities for the commutator are verified directly inside the associative algebra $\operatorname{End}_F(\mathfrak h)$. Finally, the Jacobi identity in $\mathfrak h$ shows that each inner map $\operatorname{ad}_x$ satisfies the derivation rule. [/proofplan]

[step:Show that derivations form an $F$-vector space] Let $\operatorname{End}_F(\mathfrak h)$ denote the $F$-[vector space](/page/Vector%20Space) of all $F$-linear maps $\mathfrak h \to \mathfrak h$. The zero map $0_{\mathfrak h}:\mathfrak h \to \mathfrak h$ satisfies \begin{align*} 0_{\mathfrak h}([u,v]_{\mathfrak h}) = [0_{\mathfrak h}(u),v]_{\mathfrak h} + [u,0_{\mathfrak h}(v)]_{\mathfrak h} \end{align*} for all $u,v \in \mathfrak h$, so $0_{\mathfrak h}\in \operatorname{Der}(\mathfrak h)$. Let $D_1,D_2 \in \operatorname{Der}(\mathfrak h)$ and let $a,b \in F$. Define the $F$-[linear map](/page/Linear%20Map) \begin{align*} D:\mathfrak h &\to \mathfrak h \\ u &\mapsto aD_1(u)+bD_2(u). \end{align*} For all $u,v \in \mathfrak h$, using the derivation identities for $D_1$ and $D_2$ and the bilinearity of $[\cdot,\cdot]_{\mathfrak h}$, \begin{align*} D([u,v]_{\mathfrak h}) &= aD_1([u,v]_{\mathfrak h}) + bD_2([u,v]_{\mathfrak h}) \\ &= a\bigl([D_1(u),v]_{\mathfrak h} + [u,D_1(v)]_{\mathfrak h}\bigr) + b\bigl([D_2(u),v]_{\mathfrak h} + [u,D_2(v)]_{\mathfrak h}\bigr) \\ &= [aD_1(u)+bD_2(u),v]_{\mathfrak h} + [u,aD_1(v)+bD_2(v)]_{\mathfrak h} \\ &= [D(u),v]_{\mathfrak h} + [u,D(v)]_{\mathfrak h}. \end{align*} Thus $D\in \operatorname{Der}(\mathfrak h)$, and $\operatorname{Der}(\mathfrak h)$ is an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$. [/step]

[step:Compute the derivation rule for the commutator]Let $D_1,D_2 \in \operatorname{Der}(\mathfrak h)$. Define the commutator map \begin{align*} C:\mathfrak h &\to \mathfrak h \\ u &\mapsto D_1(D_2(u))-D_2(D_1(u)). \end{align*} Since $D_1$ and $D_2$ are $F$-linear, $C$ is $F$-linear. Fix $u,v \in \mathfrak h$. First apply $D_2$ to $[u,v]_{\mathfrak h}$ and then apply $D_1$: \begin{align*} D_1(D_2([u,v]_{\mathfrak h})) &= D_1\bigl([D_2(u),v]_{\mathfrak h} + [u,D_2(v)]_{\mathfrak h}\bigr) \\ &= [D_1(D_2(u)),v]_{\mathfrak h} + [D_2(u),D_1(v)]_{\mathfrak h} \\ &\quad + [D_1(u),D_2(v)]_{\mathfrak h} + [u,D_1(D_2(v))]_{\mathfrak h}. \end{align*} Similarly, \begin{align*} D_2(D_1([u,v]_{\mathfrak h})) &= [D_2(D_1(u)),v]_{\mathfrak h} + [D_1(u),D_2(v)]_{\mathfrak h} \\ &\quad + [D_2(u),D_1(v)]_{\mathfrak h} + [u,D_2(D_1(v))]_{\mathfrak h}. \end{align*} Subtracting the second identity from the first cancels the two mixed terms and gives \begin{align*} C([u,v]_{\mathfrak h}) &= [C(u),v]_{\mathfrak h} + [u,C(v)]_{\mathfrak h}. \end{align*} Therefore $C\in \operatorname{Der}(\mathfrak h)$. Since $C=[D_1,D_2]_{\operatorname{Der}}$, the commutator bracket is closed on $\operatorname{Der}(\mathfrak h)$.[/step]

[guided]The only point needing computation is closure under the commutator. We define the candidate commutator as the $F$-linear map \begin{align*} C:\mathfrak h &\to \mathfrak h \\ u &\mapsto D_1(D_2(u))-D_2(D_1(u)). \end{align*} To prove $C$ is a derivation, we must prove \begin{align*} C([u,v]_{\mathfrak h}) = [C(u),v]_{\mathfrak h}+[u,C(v)]_{\mathfrak h} \end{align*} for arbitrary $u,v \in \mathfrak h$. Start with the first composition. Since $D_2$ is a derivation, \begin{align*} D_2([u,v]_{\mathfrak h}) = [D_2(u),v]_{\mathfrak h}+[u,D_2(v)]_{\mathfrak h}. \end{align*} Now apply $D_1$ to both bracket terms, using both $F$-linearity and the derivation rule for $D_1$: \begin{align*} D_1(D_2([u,v]_{\mathfrak h})) &= D_1\bigl([D_2(u),v]_{\mathfrak h}+[u,D_2(v)]_{\mathfrak h}\bigr) \\ &= [D_1(D_2(u)),v]_{\mathfrak h} + [D_2(u),D_1(v)]_{\mathfrak h} \\ &\quad + [D_1(u),D_2(v)]_{\mathfrak h} + [u,D_1(D_2(v))]_{\mathfrak h}. \end{align*} Interchanging the roles of $D_1$ and $D_2$ gives \begin{align*} D_2(D_1([u,v]_{\mathfrak h})) &= [D_2(D_1(u)),v]_{\mathfrak h} + [D_1(u),D_2(v)]_{\mathfrak h} \\ &\quad + [D_2(u),D_1(v)]_{\mathfrak h} + [u,D_2(D_1(v))]_{\mathfrak h}. \end{align*} The reason the commutator works is that the two mixed terms appear with the same sign in both expansions: \begin{align*} [D_2(u),D_1(v)]_{\mathfrak h} \quad\text{and}\quad [D_1(u),D_2(v)]_{\mathfrak h}. \end{align*} They therefore cancel after subtraction. Hence \begin{align*} C([u,v]_{\mathfrak h}) &= D_1(D_2([u,v]_{\mathfrak h}))-D_2(D_1([u,v]_{\mathfrak h})) \\ &= [D_1(D_2(u))-D_2(D_1(u)),v]_{\mathfrak h} + [u,D_1(D_2(v))-D_2(D_1(v))]_{\mathfrak h} \\ &= [C(u),v]_{\mathfrak h}+[u,C(v)]_{\mathfrak h}. \end{align*} Thus the commutator of two derivations is again a derivation.[/guided]

[step:Verify the Lie algebra identities for the commutator] The bracket $[\cdot,\cdot]_{\operatorname{Der}}$ is bilinear because composition in $\operatorname{End}_F(\mathfrak h)$ is bilinear and subtraction is bilinear. For $D \in \operatorname{Der}(\mathfrak h)$, \begin{align*} [D,D]_{\operatorname{Der}} = D\circ D-D\circ D = 0_{\mathfrak h}, \end{align*} so the bracket is alternating. Let $D_1,D_2,D_3 \in \operatorname{Der}(\mathfrak h)$. Expanding the three commutators inside $\operatorname{End}_F(\mathfrak h)$ gives \begin{align*} [[D_1,D_2]_{\operatorname{Der}},D_3]_{\operatorname{Der}} &= D_1D_2D_3-D_2D_1D_3-D_3D_1D_2+D_3D_2D_1, \\ [[D_2,D_3]_{\operatorname{Der}},D_1]_{\operatorname{Der}} &= D_2D_3D_1-D_3D_2D_1-D_1D_2D_3+D_1D_3D_2, \\ [[D_3,D_1]_{\operatorname{Der}},D_2]_{\operatorname{Der}} &= D_3D_1D_2-D_1D_3D_2-D_2D_3D_1+D_2D_1D_3. \end{align*} Here juxtaposition denotes composition of maps. Adding these identities, every triple composition occurs once with coefficient $1$ and once with coefficient $-1$. Hence \begin{align*} [[D_1,D_2]_{\operatorname{Der}},D_3]_{\operatorname{Der}} + [[D_2,D_3]_{\operatorname{Der}},D_1]_{\operatorname{Der}} + [[D_3,D_1]_{\operatorname{Der}},D_2]_{\operatorname{Der}} = 0_{\mathfrak h}. \end{align*} Thus the Jacobi identity holds. Therefore $\operatorname{Der}(\mathfrak h)$ is a Lie algebra over $F$ under the commutator bracket. [/step]

[step:Use the Jacobi identity to prove that inner adjoint maps are derivations] Fix $x \in \mathfrak h$. Define \begin{align*} \operatorname{ad}_x:\mathfrak h &\to \mathfrak h \\ y &\mapsto [x,y]_{\mathfrak h}. \end{align*} The map $\operatorname{ad}_x$ is $F$-linear because the bracket $[\cdot,\cdot]_{\mathfrak h}$ is bilinear. Let $y,z \in \mathfrak h$. The Jacobi identity in $\mathfrak h$ gives \begin{align*} [x,[y,z]_{\mathfrak h}]_{\mathfrak h} + [y,[z,x]_{\mathfrak h}]_{\mathfrak h} + [z,[x,y]_{\mathfrak h}]_{\mathfrak h} = 0. \end{align*} Using skew-symmetry of the bracket, \begin{align*} [y,[z,x]_{\mathfrak h}]_{\mathfrak h} = -[y,[x,z]_{\mathfrak h}]_{\mathfrak h}, \qquad [z,[x,y]_{\mathfrak h}]_{\mathfrak h} = -[[x,y]_{\mathfrak h},z]_{\mathfrak h}. \end{align*} Substituting these identities into Jacobi and rearranging yields \begin{align*} [x,[y,z]_{\mathfrak h}]_{\mathfrak h} = [[x,y]_{\mathfrak h},z]_{\mathfrak h} + [y,[x,z]_{\mathfrak h}]_{\mathfrak h}. \end{align*} In terms of $\operatorname{ad}_x$, this is \begin{align*} \operatorname{ad}_x([y,z]_{\mathfrak h}) = [\operatorname{ad}_x(y),z]_{\mathfrak h} + [y,\operatorname{ad}_x(z)]_{\mathfrak h}. \end{align*} Since $y,z \in \mathfrak h$ were arbitrary, $\operatorname{ad}_x \in \operatorname{Der}(\mathfrak h)$. This proves the final assertion. [/step]