2
$\begingroup$

Again, I have a question ! Let $E = \mathcal{C}([0;1])$ with the $||.||_{\infty}$ norm, and let $S : E \rightarrow E$ defined by : $S(u)(x) = \int_{0}^{x}u(t)\mathrm{dt} \quad \forall u \in E$.

I have already shown that $S$ is injective but not surjective, and for all $(f_n)_n \in E^{\mathbb{N}}$ a collection such that : $\exists M>0 \; \forall n \in \mathbb{N} \quad ||f_n||_\infty \leq M$, then it exists a subseries $(S(f_{\phi(n)})_n)$ which converges on $E$.

Now, I would like to find all the $\lambda \in \mathbb{R}$ such that : $S-\lambda Id$ is a bijection. I already find that wathever $\lambda \in \mathbb{R}$, the map is injective. So, I have to find for which $\lambda$ it's surjective.

So, let $v \in E$ such that : $v(x) = (S-\lambda Id)(u)(x)$ for some $u \in E$. Then : $v(x)+\lambda u(x) = \int_{0}^{x}u(t)\mathrm{dt}$ for all $x \in [0;1]$. If I suppose $u,v \in C^1([0;1])$ for example, I have the relation : $u(x) - \lambda u'(x) = v'(x)$ and $u(0) = -\frac{1}{\lambda}v(0)$ ($\lambda \neq 0$ cause $S$ is not surjective).

And then, I have to find all the $\lambda \in \mathbb{R}$ such as for all $v \in C^1([0;1])$, this equation has a solution in $E$.

My first problem is that I don't figure out how to find all those $\lambda \in \mathbb{R}$. I think it's okay for all $\lambda \in \mathbb{R^{*}}$ (cause it's a linear differential equation or the first order), but it seems me weird.

And this is my second problem : I tried to show that if $S-\lambda Id$ is a surjective from $C^1([0;1])$ to $C^1([0;1])$, then it's surjective from $E$ to $E$. To do that, I wanted to use the fact that : $\overline{C^1([0;1])} = C([0;1])$ and the fact that $S-\lambda Id$ is continuous.

So, let $u \in E$ and $(u_n)_n \in (C^1)^{\mathbb{N}}$ such that : $u_n \rightarrow u$. Let $(v_n)_n \in (C^1)^{\mathbb{N}}$ such that : $\forall n \in \mathbb{N} \quad (S-\lambda Id)(v_n) = u_n$. As $(u_n)_n$ has a subseries which converges (cause the series is bounded), if I have $(S(v_n))_n$ which admit as well a subseries which converges, then by the continuity of $S-\lambda Id$ I can conclude.

But I'm stuck, cause I don't have any argument to say that $(S(v_n))_n$ has a subseries which converges, and so, I can't conclude. Though, it would have just be enough that $(v_n)_n$ was bounded (for $||.||_\infty$), but why would it be ?

Any help would be appreciated ! :)

Thank you !

$\endgroup$
  • $\begingroup$ I think the map is surjective for any $\lambda \neq 0$. Are you familiar with integrating factors? You can solve the equation explicitly for $u$ in terms of $v$ by using an integrating factor. $\endgroup$ – Kavi Rama Murthy Sep 15 '18 at 23:46
  • $\begingroup$ I tried to do that, and I find a solution, but first I have to suppose $v$ differentiable, and then, if I do so and I solve the equation, the solution is depending of $v, \lambda$, so it seems okay, but to check if the $u$ I find for the solution of the EDO is also the $u$ which is solution of $S-\lambda Id(u) = v$, I have to suppose again $v$ differentiable. $\endgroup$ – ChocoSavour Sep 16 '18 at 9:05
  • $\begingroup$ Actually, I think it's okay. I use your method in order to prove that the map is surjective on $C^1([0;1])$, and then I solve the equation using integrating factor as you suggest, which permits me to prove that for any $u \in E$, if $(u_n)_n \in (C^1)^{\mathbb{N}}$ is such that $u_n \rightarrow u$ and $(v_n)_n \in E^{\mathbb{N}}$ such that $S-\lambda Id(v_n) = u_n$ for all n, then $(v_n)_n$ is bounded for $||.||\infty$ and it permits me to find a $(S(v_{\phi (n)}))_n$ which converges, and so a $(v_{\phi (n)})_n$ which converges and finally to conclude. Thank you ! $\endgroup$ – ChocoSavour Sep 16 '18 at 9:55
  • $\begingroup$ Welcome! Glad that you could complete the argument. $\endgroup$ – Kavi Rama Murthy Sep 16 '18 at 11:39
2
$\begingroup$

Notice that

\begin{align} \|S^2f\|_\infty &= \sup_{y \in [0,1]}\left|\int_0^y \int_0^x f(t)\,dt\,dx\right|\\ &\le \sup_{y \in [0,1]}\int_0^y \int_0^x |f(t)|\,dt\,dx\\ &\le \|f\|_\infty \sup_{y \in [0,1]}\int_0^y \int_0^x 1\,dt\,dx\\ &= \|f\|_\infty \int_0^1 x\,dx\\ &= \frac12\|f\|_\infty \end{align} so $\|S^2\|\le \frac12$. Inductively you can show that $\|S^n\| \le \frac1{n!}$ so spectral radius of $S$ is

$$r(S) = \lim_{n\to\infty} \|S^n\|^{1/n} \le \lim_{n\to\infty} \frac1{\sqrt[n]{n!} } = 0$$

Therefore the spectrum of $S$ is simply $\sigma(S) = \{0\}$ so for any scalar $\lambda \ne 0$ the map $S - \lambda I$ is bijective.

$\endgroup$
  • 1
    $\begingroup$ For students who have not heard of "spectral radius", these calculations can also be used to solve the problem. If $Sf = \lambda f$ for some nonzero $f$, we wena to show $\lambda = 0$. The first calculation here shows $|\lambda^2| \le \frac{1}{2}$. And the further calculations show $|\lambda|^n \le \frac{1}{n!}$. And therefore $\lambda = 0$. $\endgroup$ – GEdgar Sep 16 '18 at 13:06
  • $\begingroup$ Thank you for the explanations ! $\endgroup$ – ChocoSavour Sep 16 '18 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.