Proving an operator is compact.

Question

I'm studying functional analysis and I ask this question mainly because I want a feedback on my proof and suggestions to formalize better my argument (if not already well formalized); any other hint to work on this kind of problem in a better way is well accepted and desired.

The exercise I was facing is the following:

Let $T$ be the operator $$T \colon C^0([0,1],\mathbb{R},||\cdot||_{\infty}) \to C^0([0,1],\mathbb{R},||\cdot||_{\infty})$$ defined by, $\forall u \in C^0([0,1],\mathbb{R})$ (which from now on I'll denote by $X$), $$(Tu)(x) := \int_0^x \sqrt t \cdot u(t)dt.$$ Prove that the operator is compact and compute the point spectrum $\sigma_p(T)$ and the full spectrum $\sigma(T)$.

I'll write now my solution divided in parts.

1. $T$ is linear. The proof is obvious.
2. $T$ is continuous. To prove this, since it's linear and $(C^0([0,1],\mathbb{R},||\cdot||_{\infty})$ is a Banach space, I just need to prove that it's bounded. For any $x \in X$ we have:

$$|Tu(x)| = \left| \int_0^x \sqrt t \cdot u(t)dt \right|\leq \int_0^1 |\sqrt t \cdot u(t)dt| \leq ||u||_{\infty}\int_0^1 \sqrt t dt = \frac{2}{3}||u||_{\infty} < \infty$$ where the second inequality follows from Cauchy-Schwarz and the last one follows from the fact that $u$ is bounded since it's continuous on a compact interval. By taking the supremum we have $$||Tu||_{\infty} = \sup_{x \in X}|Tu(x)| \leq \frac{2}{3} ||u||_{\infty}$$ and so the operator is bounded and hence continuous.

3. We want now to prove that T is compact i.e. for every bounded sequence $(u_n)_n$, $||u_n||_{\infty} \leq M$ for some $M > 0$ and $\forall n \in \mathbb{N}$, the image $(Tu_n)_n$ has a convergent subsequence. By using Ascoli-Arzelà theorem this is the same as proving $(Tu_n)_n$ is uniformly equicontinuous and uniformly bounded. Now, $\forall x \in [0,1]$, $$|(Tu_n)(x)| \leq \int_0^1 |\sqrt t| \cdot |u(t)| dt \leq \frac{2}{3}M$$ and by taking the supremum we have $$||Tu_n||_{\infty} \leq \frac{2}{3}M$$ and so $(Tu_n)$ is uniformly bounded. Let's now prove it's uniformly equicontinuous. We know $\sqrt t$ is continuous in $X$ since it's constant with respect to the variable $x$. Let $\delta$ such that $|x - \tilde{x}| < \delta $. We have: $$|(Tu_n)(x) - (Tu_n)(\tilde{x})| = \left| \int_x^{\tilde{x}} \sqrt t \cdot u(t)dt \right| \leq M \cdot \delta$$ and so by taking $\delta = \frac{\epsilon}{M}$ we have the uniform equicontinuity.

4. Since $T$ is compact we have $\sigma(T) = \sigma_p(T) \cup \{0\}$. Let $\lambda \in \sigma_p(T)$ an eigenvalue with eigenvector $u$. Then: $$(Tu)(x) = \int_0^x \sqrt t \cdot u(t) dt = \lambda u(x)$$ and differentiating both sides we obtain $$\sqrt x \cdot u(x) = \lambda u'(x)$$ with $u(0)=0$. The Cauchy's problem: \begin{equation} \left\{\begin{aligned} u'(x) &= \frac{\sqrt x \cdot u(x)}{\lambda} \\ u(0) &= 0 \end{aligned} \right. \end{equation} has $u \equiv 0$ as the only solution for Gronwall's Lemma since $\int_0^1 \frac{\sqrt x}{\lambda}$ is finite. In conclusion we have $\sigma_p(T) = \emptyset$ and so $\sigma(T) = \{0\}$.

So, here it was my proof. I'll tell you quickly where I suspect I may have done something imperfect and that's why I'm here to ask.

1. Chosing $\delta$ as I needed was fair? Or do I need to do something else to prove the equicontinuity?
2. Is using Gronwall's Lemma for the computation of the eigenvalue like shooting with a cannon to a fly? Is there a simpler method I should use in a first course of functional analysis without calling to differential equations?

I thank in advance any of you willing to help me correct/confirm this proof and clear my mind from doubts.

Answer 1

1. Well, I guess you also have to prove $T$ is well-defined, i.e. $T(u)$ is continuous, though also trivial (actually overdone when proving $T(u_n)$ are uniformly equicontinuous in the next.)

2. There is a typo $|\sqrt{t}\cdot u(t)dt|$ should be $|\sqrt{t} \cdot u(t)|dt$. The second inequality is not Cauchy-Schwartz, but simply $f\le g\Rightarrow \int f \le \int g$.

3. Don't repeat the computation in the last step. $\|T(u_n)\|\le \|T\|\|u_n\|\le \|T\|M$, or simply a bounded operator sends a bounded set to a bounded one. It's OK to pick $\delta$ the way you did. To be slightly more clear, $|\sqrt{t} \cdot u_n(t)|= |\sqrt{t}|\cdot |u_n(t)|\le 1\cdot M = M$, hence $\int_x^{\tilde{x}}|\sqrt{t} \cdot u_n(t)|dt\le M|\tilde{x}-x|\le M\delta$.

4. Gronwall is a very good tool for this. Another one would be to notice $f(x,u)=\frac{\sqrt{x}}{\lambda} \cdot u$ is continuous and Lipschitz for $u$, hence we can apply the Picard–Lindelöf theorem to conclude the only solution is zero (but this is not as good as Gronwall, since the latter is global while Picard is local and hence needs more care if complete rigor is needed.)

Answer 2

This is an alternative proof omitting the Ascoli-Arzela theorem. For any $u\in C^0[0,1]$ let $u_n$ be the function such that $u_n(k/n)=u(k/n),$ $k=0,1,\ldots, n$ and $u_n$ is linear between $(k-1)n$ and $k/n.$ Moreover if $u$ satisfies the Lipschitz condition $$|g(x_1)-g(x_2)|\le |x_1-x_2|\quad (*)$$ so does $u_n.$ Define the operator $T_n$ by the formula $T_nf=(Tf)_n.$ The operator $T_n$ is bounded since $\|u_n\|_\infty\le \|u\|_\infty$ and $T$ is bounded. The operator $T_n$ is finite dimensional operator, hence it is compact. We will show that $\|T_n-T\|\to 0,$ which would imply the compactness of $T.$ For $\|f\|_\infty\le 1$ the functions $Tf$ and $T_nf$ satisfy the Lipschitz condition $(*)$ as $|(Tf)'|\le 1.$ For fixed $x$ we have ${k-1\over n}\le x\le {k\over n}.$ Hence $$|(T_nf)(x)-(Tf)(x)| \le |(T_nf)(x)-(T_nf)(k/n)|\\ +|(Tf)_n(k/n)-(Tf)(k/n)|+|(Tf)(k/n)-(Tf)(x)|\\ \le {2\over n}+|(Tf)_n(k/n)-(Tf)(k/n)|={2\over n}$$ Thus $\|T_n-T\|_\infty \le 2/n.$

Let $V$ denote the Volterra operator $$(Vf)(x)=\int\limits_0^xf(t)\,dt $$ Then $$\|Tf\|_\infty \le \|T|f|\|_\infty\le \|V|f|\|_\infty $$ Therefore $$\|T^kf\|_\infty \le \|V^k|f|\|_\infty$$ Thus $\|T^k\|\le \|V^k\|.$ It can be proved by induction that $$(V^k)(f)(x)={1\over (k-1)!}\int\limits_0^x(x-t)^{k-1}f(t)\,dt $$ Thus $\|V^k\|\le {1\over (k-1)!}.$ We get $$\|T^k\|^{1/k}\le \|V^k\|^{1/k}\le {1\over ((k-1)!)^{1/k}}\to 0$$ Hence the spectral radius of $T$ is equal $0,$ i.e. $\sigma(T)=\{0\}.$