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Let $f \in \mathcal{H}$ be a function $f: \mathbb{R} \rightarrow \mathbb{C}$ in a Hilbert Space $\mathcal{H}$. Now suppose that for some linear operator $T$ that acts on $\mathcal{H}$ and for some $f \in \mathcal{H}$: $$T(f) = \lambda f$$ For some $\lambda \in \mathbb{C}$. That is, $f$ is an eigenfunction of $T$ with eigenvalue $\lambda$. Now suppose that $f$ can be written as $gh$, where $g,h \in \mathcal{H}$ and $$T(g) = \alpha g$$ And $$T(h) = \beta h$$ Then how do we define the operation $T(hg)$ in a way we can write it in terms of $\alpha, \beta$?

My professor said that, in this case, $\lambda = \alpha + \beta$, but this doesn’t look clear to me. I have tought that maybe the operation is $$T(gh) = T(g)h + gT(h)$$ Which would give the correct result, but I can’t see this working for some operators (ex: second order differential operator, $D_x^2$ or taking the antiderivative).

How do we define $T(gh)$ and how to relate $\lambda$ to $\alpha$ and $\beta$?

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    $\begingroup$ If we only know $T$ is linear, we cannot deduce anything specific for the connection between $\lambda$ and $\alpha,\beta$. Nevertheless, if $T$ satisfies the Leinitz rule, we get a clear connection $\lambda=\alpha+\beta$ - as you noted. If, on the other hand, $T$ preserves multiplication ($T(gh)=T(g)T(h)$) then we obviously get $\lambda=\alpha\beta$. $\endgroup$ – Berci Jun 22 '18 at 23:41
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    $\begingroup$ It is interesting to point that out. It seem fairly clear to me now that one must specify other properties of the operator in order to properly understand the relations between eigenvalues. I personally think that your comment could be turned into an answer. $\endgroup$ – Vitor C Goergen Jun 23 '18 at 20:36

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