Spectrum of the bi-Laplacian I have the following operator:
$$\begin{align*}D(A)&=\{f \in C[0,1]:f''''\in C[0,1] \text{ and } f^{(k)}(0)=0 \ \forall\ 0\leq k\leq 3\}\\Af&=-f''''.\end{align*}$$
I'm trying to find the spectrum of $A.$
My attempt: Solving the equation $(\lambda - A) g =f$ with the given boundary condition gives:
$$g(x)=\frac{1}{4^{1/4}\lambda^{3/4}}\int_0^x f(y) \left(\sin (c_{\lambda}(x-y))\cosh (c_{\lambda}(x-y))-\cos (c_{\lambda}(x-y))\sinh (c_{\lambda}(x-y)) \right) \ dy$$
where $c_{\lambda}=\left(\frac{\lambda}{4}\right)^{1/4}.$
I'm unsure how to proceed from here. I think from the expression of $g,$ we must have $0\in \sigma(A)$ but I am not so sure since the integrand also vanishes at $\lambda=0.$
How do I proceed from here? 
 A: By the formula you derived I would have also expected that $0 \in \sigma(A)$. So I am quite curious if you didn't make a mistake somewhere. But even if your formula holds I think you can show that $0 \in \rho(A)$ as follows:
Your formula implies that $\sigma(A) \subseteq \{0\}$ since the resolvent is well-defined and bounded by the formula for each $\lambda \neq 0$. We assume that $0 \in \sigma(A)$. Then $0$ is obviously contained in the topological boundary of the spectrum and since $A$ is a closed operator (I would assume just by experience, haven't checked it though) that means that $0$ is contained in the approximate point spectrum. Hence, there exists a sequence $(f_n)_{n \in \mathbb N}$ in $D(A)$ with $\lVert f_n \rVert_\infty = 1$ and $\lim_{n \to \infty} \lVert Af_n \rVert_\infty = \lim_{n \to \infty} \lVert {f_n}'''' \rVert_\infty = 0.$
Now lets see that this is absurd: As a result of the mean value theorem one has the estimate
$$ \lvert f(x) - f(y) \rvert \leq \lVert f' \rVert_\infty \lvert x - y \rvert$$
for all $x, y \in [0,1]$ and $f \in C^1[0,1]$. For $f \in D(A)$ one even has
$$\lvert f(x) \rvert = \lvert f(x) - f(0) \rvert \leq \lVert f' \rVert_\infty \lvert x - 0\rvert \leq \lVert f' \rVert_\infty$$
and therefore $\lVert f \rVert_\infty \leq \lVert f' \rVert_\infty$ by passing to the supremum. Now due to the boundary conditions you can iterate this argument to obtain $\lVert f \rVert_\infty \leq \lVert f'''' \rVert_\infty = \lVert Af \rVert_\infty$ for all $f \in D(A)$. 
So finding a sequence $(f_n)_{n \in \mathbb N}$ in $D(A)$ with $\lVert f_n \rVert_\infty = 1$ and $\lim_{n \to \infty} \lVert Af_n \rVert_\infty = 0$ is absurd. Hence, we infer that $0 \not \in \sigma(A)$. 
As I said I didn't check that your formula above is valid. But from experience I would assume that $\sigma(A) = \emptyset$. So maybe check your formula.
