# Find linear operator given set of eigenvalues

The condensed problem:

I have a bounded, compact, self-adjoint, linear operator $A$ on $L^2([a,b];\mathbb{R})$ with positive eigenvalues $\{\frac1{\lambda_i}\}_i$.

Let $\lambda > 0$. Is there an operator $\tilde{A}$ with eigenvalues $\{\frac1{\lambda_i + \lambda}\}_i$? Does it have the same eigenfunctions? Is it bounded, compact, self-adjoint and linear?

My actual problem (maybe the details are helpful):

I have $A \phi = \int_a^b G(|t-s|) \phi(s) ds$ on $L^2([a,b];\mathbb{R})$, where $G\colon[0,T]\to \mathbb{R}$ is non-vanishing, continuous and positive definite, i.e. $$\int_a^b \int_a^b G(|t-s|) \phi(t) \phi(s) ds dt > 0$$ for all $\phi \neq 0$. By a small extension of Mercer's theorem, all eigenvalues $\{\frac1{\lambda_i}\}_i$ are positive, and the corresponding eigenfunctions $\{\phi_i\}_i$ are continuous.

I have expressions of the form $$\sum_i \frac1{\lambda_i + \lambda} \langle \phi, \phi_i\rangle \phi_i$$ and would like to conclude with the Hilbert-Schmidt theorem that these are equal to $\tilde{A} \phi$ for some nice operator $\tilde{A}$.

• You can assume that the eigenfunctions form an orthonormal family. Define $\tilde{A}\phi_i := \frac{1}{\lambda_i + \lambda}\phi_i$. This defines - by linear extension and then continuous extension - a positive self-adjoint compact operator on the closed subspace spanned by the $\phi_i$. If that's all of $L^2$, done. Otherwise, on the orthognoal complement set $\tilde{A} = A$. – Daniel Fischer Nov 4 '16 at 19:43
• Hooray, thanks! Can I say anything more about the new operator? – Elias Strehle Nov 4 '16 at 21:31
• What more do you want? One can write $\tilde{A} = B\circ A$ with a positive self-adjoint $B$, we have $\lVert \tilde{A}\rVert < \lVert A\rVert$, by construction $\tilde{A}$ has exactly the same eigenspaces as $A$ has. Actually, since $A$ is self-adjoint, the closed span of the eigenfunctions of $A$ is the whole space, so we can forget the last bit of my first comment. – Daniel Fischer Nov 4 '16 at 21:48