Let $V : L^2(0,1) \to L^2(0,1)$ be given as follows $$Vu(x)=\int_0^x{u(t) dt}$$

We know that $\sigma(V)=\{0\}$. How to find $(V-\lambda I)^{-1}$ when $\lambda \neq 0$?

I tried to find it at least for continuous functions, but even then I get stucked in calculations.


An explicit expression for $(V - \lambda I)^{-1}u$ can be given by the theory of Laplace transformation. We observe that $$ (1\star u)(x) = (Vu)(x)$$ and thus $$ \mathcal{L}[(V-\lambda I)u](s) = \left(s^{-1} - \lambda\right) U(s),$$ with $U(s)$ the Laplace transform of $u(x)$.

The inverse is thus given by $$ \mathcal{L}[(V-\lambda I)^{-1} u](s) = \frac{1}{s^{-1} -\lambda} U(s)$$ such that $$ \mathcal{L}[(V-\lambda I)^{-1} (V-\lambda I)u](s) = \frac{1}{s^{-1} -\lambda} \mathcal{L}[(V-\lambda I)u](s) = U(s).$$

Using the inverse Laplace transform, we obtain $$ (V-\lambda I)^{-1} u(x) = \mathcal{L}^{-1} \left[ \frac{1}{s^{-1} -\lambda} U\right](x) = \left(\mathcal{L}^{-1}\frac{1}{s^{-1} -\lambda} \right)\star u(x).$$

We obtain $$ \mathcal{L}^{-1}\frac{1}{s^{-1} -\lambda} = -\lambda^{-1} \delta - \lambda^{-2}e^{x/\lambda} $$ and thus $$(V-\lambda I)^{-1} u(x) = -\lambda^{-1} u(x) -\lambda^{-2} \int_0^x \!dy\,e^{(x-y)/\lambda} u(y). $$


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