Derivations of semisimple Lie algebras are inner

On $\mathrm{Der}(\mathfrak{g})$ define $B(D,E)={\mathrm{tr}}_{\mathfrak{g}}(DE)$. By [L2] this trace form is invariant. Its restriction to the ideal $\mathrm{ad}(\mathfrak{g})$ is the Killing form, which is nondegenerate by [L3]. Finite-dimensional linear algebra therefore gives $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}(\mathfrak{g})\oplus \mathrm{ad}(\mathfrak{g}{)}^{\perp }$, and [L5] makes the orthogonal complement an ideal.

The two ideals in step 1.1 commute: their bracket lies in their intersection, which is zero. If $D$ is in the orthogonal complement, then $0=[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{D(x)}$ for every $x$; the last equality is the derivation identity. By [L4], $D(x)=0$ for every $x$, so $D=0$. Thus every derivation is inner.

If ${\mathrm{ad}}_x={\mathrm{ad}}_y$, then $x-y$ is central and [L4] gives $x=y$. For $\mathfrak{g}=0$, both derivation algebras are zero and uniqueness is vacuous.