[EX] An integral equation

[gugo82]

Here's a nice Functional Analysis exercise (I answered on another Math forum) to which I added some extra questions.

***
Few prerequisites:

Testo nascosto, fai click qui per vederlo

As usual let:
\[
L^2(0,1):=\left\{ u:[0,1]\to \mathbb{R}|\ u\text{ is Lebesgue measurable and} \int_0^1u^2(t)\ \text{d} t<\infty\right\}\; ;
\]
the space \(L^2(0,1)\) is a Hilbert space with inner product:
\[
\langle u,v \rangle := \int_0^1 u(t)\ v(t)\ \text{d} t
\]
and norm \(\lVert u \rVert_2:=\left( \int_0^1u^2(t)\ \text{d} t\right)^{1/2}\).
A linear operator \(T:L^2(0,1)\to L^2(0,1)\) is said to be:

• bounded if there exists a constant \(M\in [0,\infty[\) s.t.:
  \[
  \lVert Tu\rVert_2\leq M\ \lVert u\rVert_2
  \]
  for all \(u\in L^2(0,1)\).
  
• selfadjoint iff:
  \[
  \langle Tu,v\rangle = \langle u,Tv\rangle
  \]
  forall \(u,v\in L^2(0,1)\);
  
• compact iff it transforms bounded subsets of \(L^2(0,1)\) into relatively compact subsets.

On the other hand, let:
\[
C([0,1]):= \left\{ u:[0,1]\to \mathbb{R}|\ u \text{ is continuous in } [0,1]\right\} \; ;
\]
then \(C([0,1])\) is a Banach space with norm \(\lVert u\rVert_\infty :=\max_{[0,1]} |u|\) (called \(\infty\)-norm or \(\max\)-norm).
The set \(C([0,1])\) can be regarded as a proper subspace of \(L^2(0,1)\) and, since \(\lVert u\rVert_2\leq \lVert u\rVert_\infty\) for \(u\in C([0,1])\), the subspace topology of \(C([0,1])\) is finer than its topology as a Banach space.
A linear operator \(S:C([0,1])\to C([0,1])\) is called:

• bounded iff there exists a constant \(N\in [0,\infty[\) s.t.:
  \[
  \lVert Tu\rVert_\infty \leq N\ \lVert u\rVert_\infty
  \]
  for \(u\in C([0,1])\);
  
• compact iff it transforms bounded subsets of \(C([0,1])\) into relatively compact subsets.

***

Exercise:

1. Let \(T:L^2(0,1)\to L^2(0,1)\) be defined by:
\[
Tu(x) := \int_0^1 \big( \max \{x,t\} +xt\big)\ u(t)\ \text{d} t\; .
\]
Prove that \(T\) is a bounded linear operator.
Is \(T\) selfadjoint?
Is \(T\) compact?

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Hint: The first part is trivial.
For the other questions, use the fact that the integral kernel \(K(x,y):=\max \{x,y\} +xy\) is a symmetric continuous function over the unit square \([0,1]^2\).

2. Prove that the restriction of \(T\) to \(C([0,1])\) is a bounded compact linear operator mapping \(C([0,1])\) into itself.
Is \(T:C([0,1]) \to C([0,1])\) surjective?

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Hint: The first part is trivial.
For the surjectivity, try to improve the regularity of \(Tu\) using the Fundamental Theorem of Calculus.

3. Study the existence of solutions in \(C([0,1])\) for the integral equation:
\[
\tag{1} u(x) =\lambda\ \int_0^1 \big( \max \{x,t\} +xt\big)\ u(t)\ \text{d} t
\]
when the parameter \(\lambda\) assumes real values.

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Hints: Show that \(u\) solves (1) iff it solves an eigenvalue problem for a second order linear ODE under Robin boundary conditions; then solve that eigenvalue problem.

4. Assume equation (1) has some \(L^2\) solution. Is it true that each solution is in fact of class \(C^2\)?

[dissonance]

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I'd like to start by pointing out that the given integral kernel is the Green's function of a Sturm Liouville operator on \([0, 1]\). Indeed, for arbitrary \(u \in L^2[0, 1]\) write \(y(x)=Tu(x)\) and evaluate

\[\tag{1} y(x)=x\int_0^x(1+\xi)u(\xi)\, d\xi + (1+x)\int_x^1\xi u(\xi)\, d\xi; \]

so \(y\) is absolutely continuous and its derivative is

\[\tag{2} y'(x)=\int_0^x u(\xi)\, d\xi + \int_0^1 \xi u(\xi)\, d\xi;\]

then, again, \(y'\) is absolutely continuous and its derivative is

\[\tag{3} y''(x)=u(x).\]

We infer from (1), (2) that \(y\) satisfies boundary conditions \(y(0)-y'(0)=0,\ y(1)-y'(1)=0\), which together with (3) means that \(y\) satisfies the following problem:

\[\tag{SL} \begin{cases} y''(x)=u(x), \quad u \in L^2[0, 1] \\ y(0)-y'(0)=0, y(1)-y'(1)=0 \end{cases}\]

[dissonance]

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1. Write \(G(x, \xi)=\max(x, \xi)+x\xi.\) Notice that \(G(x, \xi)=G(\xi, x)\) and that \(G\) is square-integrable on \([0, 1]\times [0, 1]\). A standard application of Cauchy-Schwarz inequality yields \(\lVert T \rVert_{\rm{op}}\le \lVert G\rVert_{L^2([0, 1]^2)}\), so \(G\) is bounded. Also, it is symmetric, hence self-adjoint. Now take a sequence of simple functions \(0 \le S_n(x, \xi) \uparrow G(x, \xi)\). Define \(T_nu(x)=\int d \xi S_n(x, \xi)u(\xi)\). This is a sequence of finite rank operators such that \(\lVert T_n-T\rVert_{\rm{op}}\le\lVert S_n-G\rVert_{L^2([0, 1]^2)}\to 0\). So \(T\) is compact.

2. Now consider \(T\) as an operator on \(C[0, 1]\). We have \(\lVert T \rVert_{\rm{op}}\le \int dxd\xi G(x, \xi)\), so \(T\) is bounded. Also, as we have seen in the previous post, \(Tu\) is \(C^1\) and \(\lVert (Tu)'\rVert_{\infty}\le 2\lVert u \rVert_{\infty}\). This implies that \(T\) is compact (by Ascoli-Arzelà's theorem) and that it is not surjective.

3. As we have already noticed in the previous post, the given problem can be restated in differential terms:

\[\tag{SL} \begin{cases} y''(x)=\lambda y(x) \\ y(0)-y'(0)=0, y(1)-y'(1)=0 \end{cases}\]

For \(\lambda > 0\) the general integral of this equation is \(y(t)=Ae^{\sqrt{\lambda}t}+Be^{-\sqrt{\lambda}t}\);
for \(\lambda =0\) it is \(y(t)=At+B\);
for \(\lambda<0\) it is \(y(t)=A\cos(\omega t)+B \sin(\omega t), \quad \omega^2=-\lambda.\)

As it is readily seen, we only have non-trivial solutions for \(\lambda=1\) or\(\lambda=-\pi, -2\pi, -3\pi\ldots -n\pi, \ldots\). An orthonormal system of eigenvectors is

\[u_0(t)=\frac{\sqrt{2}}{\sqrt{e^2-1}}e^t, \]
\[u_n(t)=\frac{\sqrt{2}}{\sqrt{1+(n\pi)^2}}\left[n \pi \cos(n \pi t)+\sin(n \pi t)\right]\quad n=1, 2, 3, \ldots.\]

From the general Sturm-Liouville theory, or from the spectral theory of compact operators, we know that this system is an orthonormal basis of \(L^2[0, 1]\). Notice that those eigenvalues are all simple (= of multiplicity 1).

4. Yes. A solution of (1) must be a scalar multiple of one of the eigenvectors we just found or the a.e. null function.

[gugo82]

Mostly correct... But you're missing one singular value \(\lambda\).
Find it!

My solution follows:

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1. \(T\) is defined via the kernel \(K(x,t):= \max \{x,t\}+xt\), which is a function of class \(C([0,1]^2)\) and it is also symmetric. It then follows from standard arguments that \(T\) is a (bounded) linear selfadjoint compact operator in \(L^2(0,1)\).

2. Reasoning as before we conclude that \(T\) is a compact linear operator also over \(C([0,1])\).

From the definition of \(T\) follows:
\[
\tag{I} Tu(x) =\int_0^x t\ u(t)\ \text{d} t +x\int_x^1 u(t)\ \text{d} t +x\int_0^1 t\ u(t)\ \text{d} t\; ;
\]
since \(u\) is continuous Fundamental Theorem of Integral Calculus applies yielding \(Tu\in C^1([0,1])\). Therefore \(T\) maps \(C([0,1])\) into \(C^1([0,1])\), which is a proper subspace of \(C([0,1])\), and thus \(T\) cannot be surjective.

3. The same as your solution... But with one more singular value.

4. A sort of bootstrap argument can be used.
If \(u\) solves \(u=\lambda Tu\), then \(u\in AC(0,1)\) (because it is an integral function of an element of \(L^2\)); but then it follows from (I) that \(u\) is of class \(C^1([0,1])\); using (I) again, it is easy to check that \(u\in C^2([0,1])\)... In general, identity (I) can be used to prove that \(u\in C^k\ \Rightarrow \ u\in C^{k+1}\) for each \(k\in \mathbb{N}\), hence induction yields \(u\in C^\infty ([0,1])\).

[dissonance]

Faulty computations, as usual!
Now I corrected it. By the way, that bootstrap argument is nice!

I've got one more question regarding this exercise, but I will submit it later (as soon as I've got some spare time).