 Solution

The spectrum of a bounded operator is $\sigma (T)=\{\lambda :\lambda I-T\text{ has no bounded two-sided inverse}\}$. Its point spectrum is the set of eigenvalues, where $\lambda I-T$ is not injective.

Here $T$ is linear and takes continuous functions to continuously differentiable functions. Also $\Vert Tx{\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$; the constant function one attains this bound at $t=1$, so $\Vert T\Vert =2$. Its range is contained in $C^1([0,1])$, a proper subset of $C([0,1])$, so zero is in its spectrum. It is not an eigenvalue: $Tx=0$ gives $x(0)=0$ at zero and $x=0$ upon differentiation.

For a nonzero eigenvalue, $\lambda x=Tx$ implies $x^{\prime }=x/\lambda$ and $(\lambda -1)x(0)=0$. A nonzero solution is therefore possible exactly at $\lambda =1$, with eigenfunctions $x(t)=c{e}^t$. Hence $p\sigma (T)=\{1\}$.

For the proposed resolvent putThe proposed $x$ is $y/\lambda +y(0)e^{t/\lambda }/[\lambda (\lambda -1)]+F/{\lambda }^2$. Consequently $x(0)=y(0)/(\lambda -1)$ andAdding $x(0)$ and subtracting from $\lambda x(t)$ gives exactly $y(t)$. Thus $\lambda I-T$ is onto for $\lambda \ne 0,1$. It is also injective by the eigenvalue calculation. The displayed formula defines a bounded map in the supremum norm, since its exponential kernels are bounded on the compact interval. It is therefore the bounded inverse, and