Spectrum of the operator of indefinite integration

Question

Find the spectrum of the operator $A:C_0[0,1]\to C_0[0,1]$, $Ax(t)=\displaystyle\int\limits_0^tx(\tau)d\tau$, where $C_0[0,1]$ -- is the space of continuous functions on a segment $[0,1]$ such that $x(0)=0$ with the usual $\sup$-norm.

Since the operator is compact it can only have eigenvalues plus $\{0\}$. I showed that there are no eigenvalues (include $0$), therefore it remains to explore point $0$ for membership in a continuous or residual spectrum. It's easy to see that the image of the operator consists of continuously differentiable on $[0,1]$ functions $y(t)$, such that $y(0)=y'(0)=0$. My question is how to prove or disprove that this set is dense in $C_0[0,1]$. I will be grateful for any hints.

Given a continuous function with $y(0)=0$ choose a small neighborhood of $0$ where $|y(t)|<\epsilon$. Then multiply it by a smooth monotonic step function $\phi(t)$ with all derivatives vanishing at $0$ that's $1$ outside of this neighborhood. Then $y(t)\phi(t)$ is in the image and $\sup_t|y(t)\phi(t)-y(t)|<\epsilon$. — Conifold, Jan 12 '20 at 10:31
@Conifold, thanks for your comment, but i don't understand, why $y(t)\varphi(t)$ must be differentiable? — thing, Jan 12 '20 at 10:48
You are right, I was too hasty. First, use the Weierstrass Approximation Theorem to get a polynomial such that $\sup_t|y(t)-p(t)|<\epsilon$, then use $p(t)\phi(t)$ as the approximation in the image. — Conifold, Jan 12 '20 at 11:04

score 1 · Answer 1 · answered Jan 12 '20 at 11:12

Let $n\geq2$. Choose $\varphi_n\in C^\infty$ such that $0\leq\varphi_n\leq1$ and $\varphi_n=0$ over $[0,\frac1{2n}]$ and $=1$ over $[\frac1n,1]$ and let $f_n=f\varphi_n$. Then it is easy to check that $\{f_n\}\subset R(A)$ and $$|f_n(x)-f(x)|\leq\begin{cases} 2|f(x)|\leq\frac2n\|f'\|,&x\leq \frac1n,\\ 0,&x>\frac1n. \end{cases}$$ Hence $f_n\to f$ in $C_0[0,1]$.

Spectrum of the operator of indefinite integration

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