If a Hilbert space $H$ has an ONB consisting of the eigenfunctions of an operator $A$, is $Av=\sum_{j} l_j\left<v,e_j\right>e_j$ always true?

Question

Let $H$ be a Hilbert space over the field $\mathbb{K}$ (e.g. $\mathbb{C}$). Suppose that $H$ has an orthonormal basis consisting of the eigenfunctions of a linear operator $A$ acting on $H$. Denote the ONB by $\{e_j\}_j$ and let $\{l_j\}_j$ be the eigenvalues $Ae_j = l_je_j$. We know that every $v\in H$ can be represented as $v=\sum_{j}\left<v,e_j\right>e_j$, but is it true that $Av=\sum_{j}l_j\left<v,e_j\right>e_j$? If not, what is the least amount of extra assumptions we would have to impose for $A$ for $Av=\sum_{j}l_j\left<v,e_j\right>e_j$ to hold? One lower bound of assumptions is given by the diagonalization theorem which asserts that the equality

$$Av=\sum_{j}l_j\left<v,e_j\right>e_j$$

holds when $A$ is a compact self-adjoint operator.

Answer 1

Assume $A$ is symmetric. Note that the eigenvalues are real, because $A$ is symmetric. $(e_j)$ is an orthonormal basis of $H$, so $$Av = \sum_j\langle Av,e_j \rangle e_j= \sum_j\langle v,Ae_j \rangle e_j= \sum_j\langle v,l_j e_j \rangle e_j= \sum_j l_j\langle v,e_j \rangle e_j$$

Answer 2

After several re-thinkings of this, the 'least amount of extra assumptions we would have to impose on $A$ for $$Av = \sum_{j=1}^\infty l_j \langle v, e_j \rangle e_j \quad (1)$$ to hold for every $v\in H$' is that $A$ is bounded.

Because assume $A$ is bounded. Then $$|l_j| = |l_j|\|e_j\| = \|l_j e_j\| = \|A e_j\| \le \|A\|\|e_j\| = \|A\|$$ Now let $v\in H$. Then $\sum_{j=1}^\infty |\langle v,e_j\rangle|^2 < +\infty$. Therefore $$\sum_{j=1}^\infty |l_j|^2 |\langle v,e_j\rangle|^2 \le \|A\|^2 \sum_{j=1}^\infty |\langle v,e_j\rangle|^2 < +\infty\quad (2)$$ Let $v_N = \sum_{j=1}^N \langle v, e_j\rangle e_j$, and $z_N = A(v_N) = \sum_{j=1}^N \langle v, e_j\rangle A(e_j) = \sum_{j=1}^N \lambda_j \langle v, e_j\rangle e_j$. Then $v_N\to v$, and because of $(2)$, $z_n \to z = \left(\sum_{j=1}^\infty \lambda_j \langle v, e_j\rangle e_j\right) \in H$. Since $A$ is bounded, it follows that $A(v) = z$.

Conversely, assume that $(1)$ holds for every $v\in H$. Then $\|Av\|^2 = \sum_{j=1}^\infty |l_j|^2 |\langle v, e_j\rangle|^2$. It is sufficient to show that $L = \sup_{j=1}^\infty |l_j| < +\infty$, because then $\|Av\|^2 \le L^2 \|v\|^2$ proving that $A$ is bounded. Assume by contradiction that $\{l_j\}_{j=1}^\infty$ is unbounded. Then there exists a subsequence $l_{j_k}$ such that $|l_{j_k}| > k$. Now let $w=\sum_{k=1}^\infty \frac{1}{k} e_{j_k}$. Then $\|w\|^2 = \sum_{k=1}^\infty \frac{1}{k^2} < +\infty$ so $w\in H$ but $Aw=\sum_{k=1}^\infty \frac{\lambda_{j_k}}{k} e_{j_k}$ which is a contradiction because $|\frac{\lambda_{j_k}}{k}| > 1$ for all $k$ so the series for $Aw$ cannot converge in $H$.

That said, there is a way to relax the restrictions by

• Allowing the domain of $A$ to be a subspace of $H$.
• Requiring $A$ to be closed.

In this case let $V = \{v\in H: \sum_{j=1}^\infty |\lambda_j|^2 |\langle v, e_j\rangle|^2 < +\infty\}$. If $U=\operatorname{span}\{\{e_j\}_{j=1}^\infty\}$ then

• $U\subseteq V$ because the sums are over a finite set of indices.
• For $v\in U$ equation $(1)$ holds because the sum is actually over a finite set of indices.
• For $v\in V\setminus U$ equation $(1)$ holds as well. The sum on the right hand side of $(1)$ is well-defined because of the definition of $V$, and the convergence of the right hand side to the left hand side in $(1)$ is due to the assumption that $A$ is closed. The proof is similar to the bounded case shown above and I won't repeat it.

We can further reduce the assumption on $A$ to requiring that $A$ is closable, and then use the last result on $\bar{A}$ (the closure of $A$) instead. Then $A$ is a restriction of $\bar{A}$ to some domain between $U$ and $V$. This includes the case when $A$ is symmetric, because symmetric implies closable.

If we drop the assumption that $A$ be closable then we don't really know anything about $A$ beyond the subspace $U$. It can be extended arbitrarily (by using a Hamel basis.)