We learned that if $V$ is a finite inner product space then for every linear transformation $T:V\to V$, there exists a unique linear transformation $T^*:V\to V$ such that $\forall u, v \in V: (Tv, u)=(v, T^*u)$.
The construction of $T^*$ used the fact that $V$ is finite and therefore has an orthonormal basis, which is not the case had it been infinite.
Are there infinite dimension inner product spaces such that not all linear transformations have an adjoint? Or is it somehow possible to extend this definition to infinite spaces as well?
Best Answer
This is true for (edit: bounded operators on) Hilbert spaces thanks to the Riesz representation theorem. It is false in general: let $e_1, e_2, ...$ be a sequence of orthogonal unit vectors in some infinite-dimensional Hilbert space and let $V$ be their span. Then the linear transformation
$$T(e_i) = e_1 + e_2 + ... + e_i$$
does not have an adjoint, since $\langle T(e_i), e_j \rangle = 1$ whenever $j \le i$, but $\langle e_i, T^{\ast}(e_j) \rangle$ must be $0$ for sufficiently large $i$ and fixed $j$ for any linear operator $V \to V$. This example can be modified so that $T$ is bounded.
If you are not familiar with Hilbert space theory, beware that the definition of "orthonormal basis" is different: it does not refer to a Hamel basis (which is what the word "basis" ordinarily means) but a collection of orthogonal unit vectors such that only the zero vector is orthogonal to all of them. Equivalently, it refers to a collection of orthogonal unit vectors whose span is dense (not the whole space).