Prove that $Tf(x)=\int^{1-x}_0 f(t) \, dt$ is compact and compute its spectrum.

compact-operatorsfunctional-analysisspectral-theory

Prove that $T:C^0([0,1])\longrightarrow C^0([0,1])$ defined as $$Tf(x)=\int^{1-x}_0 f(t) \, dt$$ is compact and compute its spectrum.

– Compactness:

Let $f_n$ be a bounded sequence, that is $\|f_n\|_\infty\leq C\,\,\forall\,n$. In order to prove that $Tf_n$ has a convergent subsequence we show that it is bounded and equicontinuous. It is easy to see that $T$ is continuous, then $Tf_n$ is bounded. Let $\varepsilon >0$ and $x_1, \,x_2 \in [0,\,1]$, then
$$|Tf_n(x_1)-Tf_n(x_2)|\leq \|f\|_\infty | x_1-x_2| \leq C | x_1-x_2|.$$
If we choose $\delta=\varepsilon/C$ and $| x_1-x_2| < \delta$ then
$$|Tf_n(x_1)-Tf_n(x_2)|< \varepsilon, \quad \forall \, n.$$

– Spectrum:

Since $T$ is compact it is not invertible. Then $0\in \sigma (T)$. Now I have some questions:

  • I know that for compact operators between Hilbert spaces if $\lambda\in \sigma(T),\,\lambda\neq0$ then $\lambda\in \sigma_p(T)$ and $\sigma_p(T)$ is finite or countable and $0$ is the only accumulation point of $\sigma_p(T)$. The proof that I have seen of this fact is based on the Fredholm Alternative Theorem (that I only know for Hilbert spaces). So I wonder if this property is still true if the operater is defined on a Banach space. If it is true then the exercise is more simple.

  • I tried to compute $\sigma_p$ but I had some problem. Here some attempts.
    We are interested in solve $$ Tf(x)=\int^{1-x}_0 f(t) \, dt=\lambda f(x) $$
    Since $f\in C^0$ we have $Tf\in C^1$ and $\lambda f\in C^1$ then,
    $$ -f(1-x)=\lambda f'(x) $$
    And now ? It is not possible to solve by separation of variable. How to proceed?

Thanks.

UPDATE: I have consulted Conway's book A Course in Functional Analysis and I have verified that the properties of the spectrum of a compact operator that I mentioned above are also valid in the case of Banach spaces (Theorem 7.1). Therefore it is sufficient to compute $\sigma_p(T)$ and understand if $\,0\in \sigma_c(T)\, $ or $\,0\in \sigma_r(T)$ or $\,0\in \sigma_p(T)$.

Best Answer

We have already established that $0 \in \sigma(A) \setminus \sigma_p(A)$.

Assume $Tf = \lambda f$. Then also $T^2f = \lambda^2f$.

We have

$$\lambda^2f(x) = (T^2f)(x) = \int_0^{1-x}\int_0^{1-s} f(t)\,dt\,ds$$

Taking the derivative gives:

$$\lambda^2 f'(x) = -\int_0^{1-x}f(t)\,dt$$

$$\lambda^2 f''(x) = f(1-x) = -\lambda f'(x)$$

so $$f''(x) = -\frac1{\lambda} f'(x)$$ This gives $f'(x) = Ce^{-\frac{x}\lambda}$ and then $f(x) = -\lambda C e^{-\frac{x}\lambda} + D$.

Now notice that it has to be $f(1) = 0$ so $D = \lambda C$.

Plugging this back:

$$-\lambda C e^{-\frac{x}\lambda} = -\lambda f'(x)= f(1-x) = -\lambda C e^{-\frac{1-x}\lambda} + \lambda C$$

Can you calculate $\lambda$ from here?

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