# Linear map from $\mathcal{C}([0;1])$ to $\mathcal{C}([0;1])$ and bijection

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 !

• 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. – Kavi Rama Murthy Sep 15 '18 at 23:46
• 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. – ChocoSavour Sep 16 '18 at 9:05
• 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 ! – ChocoSavour Sep 16 '18 at 9:55
• Welcome! Glad that you could complete the argument. – Kavi Rama Murthy Sep 16 '18 at 11:39

\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.
• 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 $\lambda = 0$. – GEdgar Sep 16 '18 at 13:06