# Spectral theory, Volterra integral operators and the Sturm-Liouville theorem

Jordan Bell
December 5, 2016

## 1 Banach algebras

Let $A$ be a complex Banach algebra with unit element $e$. Let $G(A)$ be the set of invertible elements of $A$. For $x\in A$, the resolvent set of $x$ is

 $\rho(x)=\{\lambda\in\mathbb{C}:\lambda e-x\in G(A)\}.$

The spectrum of $x$ is

 $\sigma(x)=\mathbb{C}\setminus\rho(x)=\{\lambda\in\mathbb{C}:\lambda e-x\not\in G% (A)\}.$

The spectral radius of $x$ is

 $r(x)=\sup\{|\lambda|:\lambda\in\sigma(x)\}.$

One proves that $\sigma(x)\subset\mathbb{C}$ is compact and nonempty and

 $r(x)=\lim_{n\to\infty}\left\|x^{n}\right\|^{1/n},$

the spectral radius formula.11 1 Walter Rudin, Functional Analysis, second ed., p. 253, Theorem 10.13. If $r(x)=0$ we say that $x$ is quasinilpotent.22 2 We say that $x\in A$ is nilpotent if there is some $n\geq 1$ such that $x^{n}=0$, and if $x$ is nilpotent then by the spectral radius formula, $x$ is quasinilpotent. $x\in A$ is quasinilpotent if and only if $\sigma(x)=\{0\}$.

###### Lemma 1.

If $x\in A$ is quasinilpotent and $|\lambda|>0$, then $S_{n}=\sum_{j=0}^{n}\lambda^{j}x^{j}\in A$ is a Cauchy sequence, and

 $(e-\lambda x)\sum_{n=0}^{\infty}\lambda^{n}x^{n}=e.$
###### Proof.

Let $0<\epsilon<|\lambda|^{-1}$. There is some $n_{\epsilon}$ such that $\left\|x^{n}\right\|^{1/n}\leq\epsilon$ for $n\geq n_{\epsilon}$. For $n>m\geq n_{\epsilon}$,

 $\left\|S_{n}-S_{m}\right\|\leq\sum_{j=m+1}^{n}|\lambda|^{j}\left\|x^{j}\right% \|\leq\sum_{j=m+1}|\lambda|^{j}\epsilon^{j},$

and because $|\lambda|\epsilon<1$, it follows that $S_{n}\in A$ is a Cauchy sequence and so converges to some $S\in A$, $S=\sum_{n=0}^{\infty}\lambda^{k}x^{k}$. Now,

 $\displaystyle(e-\lambda x)S$ $\displaystyle=(e-\lambda x)S_{n}+(e-\lambda x)(S-S_{n})$ $\displaystyle=S_{n}-\lambda xS_{n}+(e-\lambda x)(S-S_{n})$ $\displaystyle=S_{n}-\sum_{j=1}^{n+1}\lambda^{j}x^{j}+(e-\lambda x)(S-S_{n})$ $\displaystyle=e-\lambda^{n+1}x^{n+1}+(e-\lambda x)(S-S_{n}).$

Because $x$ is quasinilpotent it follows that $\left\|(e-\lambda x)S-e\right\|\to 0$. ∎

For $x\in A$ and $\lambda\in\rho(x)$, let

 $R_{x}(\lambda)=(x-\lambda e)^{-1}.$
###### Lemma 2.

If $x\in A$ is quasinilpotent and $\lambda\in\mathbb{C}$ then

 $(e-\lambda x)^{-1}=\sum_{n=0}^{\infty}\lambda^{n}x^{n}$

and if $|\lambda|>0$ then

 $R_{x}(\lambda)=-\lambda^{-1}(e-\lambda^{-1}x)^{-1}=-\lambda^{-1}\sum_{n=0}^{% \infty}\lambda^{-n}x^{n}.$

## 2 Volterra integral operators

Let $I=[0,1]$ and let $\mu$ be Lebesgue measure on $I$. $C(I)$ is a Banach space with the norm

 $\left\|f\right\|_{\infty}=\sup_{x\in I}|f(x)|,\qquad f\in C(I).$

$L^{1}(I)$ is a Banach space with the norm

 $\left\|f\right\|_{L^{1}}=\int_{I}|f(x)|dx,\qquad f\in L^{1}(I).$

For $f:I\to\mathbb{C}$, let

 $|f|_{\mathrm{Lip}}=\sup_{x,y\in I,x\neq y}\frac{|f(x)-f(y)|}{|x-y|}.$

Let $\mathrm{Lip}(I)$ be the set of those $f:I\to\mathbb{C}$ with $|f|_{\mathrm{Lip}}<\infty$. It is a fact that $\mathrm{Lip}(I)$ is a Banach space with the norm $\left\|f\right\|_{\mathrm{Lip}}=\left\|f\right\|_{\infty}+|f|_{\mathrm{Lip}}$.33 3 Walter Rudin, Real and Complex Analysis, third ed., p. 113, Exercise 11.

 $\mathrm{Lip}(I)\subset C(I)\subset L^{1}(I).$

$A=\mathscr{L}(C(I))$ is a Banach algebra with unit element $e(f)=f$ and with the operator norm:

 $\left\|T\right\|=\sup_{f\in C(I),\left\|f\right\|_{\infty}\leq 1}\left\|Tf% \right\|_{\infty},\qquad T\in A.$

For $K:I\times I\to\mathbb{C}$ and for $x,y\in I$ define

 $K_{x}(y)=K(x,y),\qquad K^{y}(x)=K(x,y).$

Let $K\in C(I\times I)$. For $f\in L^{1}(I)$ define $V_{K}f:I\to\mathbb{C}$ by

 $V_{K}f(x)=\int_{0}^{x}K(x,y)f(y)dy,\qquad x\in I.$
###### Lemma 3.

If $K\in C(I\times I)$ and $f\in C(I)$ then $V_{K}f\in C(I)$.

###### Proof.

For $x_{1},x_{2}\in I$, $x_{1}>x_{2}$,

 $\displaystyle V_{K}f(x_{1})-V_{K}f(x_{2})$ $\displaystyle=\int_{0}^{x_{1}}K(x_{1},y)f(y)dy-\int_{0}^{x_{1}}K(x_{2},y)f(y)dy$ $\displaystyle+\int_{0}^{x_{1}}K(x_{2},y)f(y)dy-\int_{0}^{x_{2}}K(x_{2},y)f(y)dy$ $\displaystyle=\int_{0}^{x_{1}}\bigg{[}K(x_{1},y)-K(x_{2},y)\bigg{]}f(y)dy+\int% _{x_{2}}^{x_{1}}K(x_{2},y)f(y)dy.$

Let $\epsilon>0$. Because $K:I\times I\to\mathbb{C}$ is uniformly continuous, there is some $\delta_{1}>0$ such that $|(x_{1},y_{1})-(x_{2},y_{2})|\leq\delta_{1}$ implies $|K(x_{1},y_{1})-K(x_{2},y_{2})|\leq\epsilon$. By the absolute continuity of the Lebesgue integral, there is some $\delta_{2}>0$ such that $\mu(E)\leq\delta_{2}$ implies $\int_{E}|f|d\mu\leq\epsilon$.44 4 http://individual.utoronto.ca/jordanbell/notes/L0.pdf, p. 8, Theorem 8. Therefore if $|x_{1}-x_{2}|<\delta=\min(\delta_{1},\delta_{2})$ then

 $\displaystyle|V_{K}f(x_{1})-V_{K}f(x_{2})|$ $\displaystyle\leq\int_{0}^{x_{1}}\epsilon|f(y)|dy+\left\|K\right\|_{\infty}% \int_{x_{2}}^{x_{1}}|f(y)|dy$ $\displaystyle\leq\epsilon\left\|f\right\|_{L^{1}}+\left\|K\right\|_{\infty}\epsilon.$

It follows that $V_{K}f:I\to\mathbb{C}$ is uniformly continuous, so $V_{K}f\in C(I)$. ∎

$\left\|V_{K}f\right\|_{\infty}\leq\left\|K\right\|_{\infty}\left\|f\right\|_{\infty}$ so $\left\|V_{K}\right\|\leq\left\|K\right\|_{\infty}$, hence $V_{K}:C(I)\to C(I)$ is a bounded linear operator, namely $V_{K}\in A$. We call $V_{K}$ a Volterra integral operator.

For $x\in I$,

 $V_{K}^{2}f(x)=\int_{0}^{x}K(x,y_{1})V_{K}f(y_{1})dy_{1}=\int_{0}^{x}K(x,y_{1})% \left(\int_{0}^{y_{1}}K(y_{1},y_{2})f(y_{2})dy_{2}\right)dy_{1}.$
 $\displaystyle V_{K}^{3}f(x)$ $\displaystyle=V_{K}^{2}V_{K}f(x)$ $\displaystyle=\int_{0}^{x}K(x,y_{1})\int_{0}^{y_{1}}K(y_{1},y_{2})V_{K}f(y_{2}% )dy_{2}dy_{1}$ $\displaystyle=\int_{0}^{x}K(x,y_{1})\int_{0}^{y_{1}}K(y_{1},y_{2})\int_{0}^{y_% {2}}K(y_{2},y_{3})f(y_{3})dy_{3}dy_{2}dy_{1}.$

For $n\geq 2$,

 $V_{K}^{n}f(x)=\int_{y_{1}=0}^{x}\int_{y_{2}=0}^{y_{1}}\cdots\int_{y_{n}=0}^{y_% {n-1}}K(x,y_{1})K(y_{1},y_{2})\cdots K(y_{n-1},y_{n})f(y_{n})dy_{n}\cdots dy_{% 1}.$

We prove that $V_{K}$ is quasinilpotent.55 5 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 53, Example 2.2.13.

###### Theorem 4.

If $K\in C(I\times I)$ then

 $\left\|V_{K}^{n}\right\|\leq\frac{\left\|K\right\|_{\infty}^{n}}{n!},$

and thus $V_{K}\in A=\mathscr{L}(C(I))$ is quasinilpotent.

###### Proof.

Let

 $\displaystyle\Phi_{n}(x)$ $\displaystyle=\int_{0}^{x}\int_{0}^{y_{1}}\cdots\int_{0}^{y_{n-1}}dy_{n}\cdots dy% _{1}$ $\displaystyle=\int_{0}^{x}\int_{0}^{y_{1}}\cdots\int_{0}^{y_{n-2}}y_{n-1}dy_{n% -1}\cdots dy_{1}$ $\displaystyle=\int_{0}^{x}\int_{0}^{y_{1}}\cdots\int_{0}^{y_{n-3}}\frac{y_{n-2% }^{2}}{2}dy_{n-2}\cdots dy_{1}$ $\displaystyle=\int_{0}^{x}\frac{y_{1}^{n-1}}{(n-1)!}dy_{1}$ $\displaystyle=\frac{x^{n}}{n!}.$

For $x\in I$,

 $\displaystyle|V_{K}^{n}f(x)|$ $\displaystyle\leq\left\|K\right\|_{\infty}^{n}\left\|f\right\|_{\infty}\int_{0% }^{x}\int_{0}^{y_{1}}\cdots\int_{0}^{y_{n-1}}dy_{n}\cdots dy_{1}$ $\displaystyle=\left\|K\right\|_{\infty}^{n}\left\|f\right\|_{\infty}\Phi_{n}(x)$ $\displaystyle=\left\|K\right\|_{\infty}^{n}\left\|f\right\|_{\infty}\frac{x^{n% }}{n!}.$

Hence

 $\left\|V_{K}^{n}\right\|\leq\frac{\left\|K\right\|_{\infty}^{n}}{n!}.$

Then

 $\left\|V_{K}^{n}\right\|^{1/n}\leq\frac{\left\|K\right\|_{\infty}}{(n!)^{1/n}}.$

Using $(n!)^{1/n}\to\infty$ we get $\left\|V_{K}^{n}\right\|^{1/n}\to 0$. Thus $V_{K}\in A$ is quasinilpotent. ∎

Theorem 4 tells us that $V_{K}$ is quasinilpotent and then Lemma 2 then tells us that for $\lambda\in\mathbb{C}$,

 $(e-\lambda V_{K})^{-1}=\sum_{n=0}^{\infty}\lambda^{n}V_{K}^{n}\in A.$ (1)

## 3 Sturm-Liouville theory

Let $Q\in C(I)$ and for $u\in C^{2}(I)$ define

 $L_{Q}u=-u^{\prime\prime}+Qu.$
###### Lemma 5.

If $u\in C^{2}(I)$ and

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1,$

then

 $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy,\qquad x\in I.$
###### Proof.

For $y\in I$, by the fundamental theorem of calculus66 6 Walter Rudin, Real and Complex Analysis, third ed., p. 149, Theorem 7.21. and using $u^{\prime}(0)=1$,

 $\int_{0}^{y}u^{\prime\prime}(t)dt=u^{\prime}(y)-u^{\prime}(0)=u^{\prime}(y)-1.$

Using $L_{Q}u=0$,

 $u^{\prime}(y)=1+\int_{0}^{y}u^{\prime\prime}(t)dt=1+\int_{0}^{y}Q(t)u(t)dt.$

For $x\in I$, by the fundamental theorem of calculus and using $u(0)=0$,

 $\int_{0}^{x}u^{\prime}(y)dy=u(x)-u(0)=u(x).$

Thus

 $\displaystyle u(x)$ $\displaystyle=\int_{0}^{x}u^{\prime}(y)dy$ $\displaystyle=\int_{0}^{x}\left(1+\int_{0}^{y}Q(t)u(t)dt\right)dy$ $\displaystyle=x+\int_{0}^{x}\left(\int_{0}^{y}Q(t)u(t)dt\right)dy.$

Applying Fubini’s theorem,

 $\displaystyle u(x)$ $\displaystyle=x+\int_{0}^{x}Q(t)u(t)\left(\int_{t}^{x}dy\right)dt$ $\displaystyle=x+\int_{0}^{x}Q(t)u(t)(x-t)dt.$

###### Lemma 6.

If $u\in C(I)$ and

 $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy,\qquad x\in I,$

then $u\in C^{2}(I)$ and

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1.$
###### Proof.
 $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy,\qquad x\in I,$

then

 $u(x)=x+x\int_{0}^{x}Q(y)u(y)dy-\int_{0}^{x}yQ(y)u(y)dy,$

and using the fundamental theorem of calculus,

 $\displaystyle u^{\prime}(x)$ $\displaystyle=1+\int_{0}^{x}Q(y)u(y)dy+xQ(x)u(x)-xQ(x)u(x)=1+\int_{0}^{x}Q(y)u% (y)dy$

hence

 $\displaystyle u^{\prime\prime}(x)$ $\displaystyle=Q(x)u(x),$

and so

 $L_{Q}u=-u^{\prime\prime}+Qu=-Qu+Qu=0.$

$u(0)=0$ and $u^{\prime}(0)=1$, so

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1.$

###### Lemma 7.

Let $Q\in C(I)$ and let $K(x,y)=(x-y)Q(y)$, $K\in C(I\times I)$. Let $u_{0}(x)=x$, $u_{0}\in C(I)$. Then $\sum_{j=0}^{n}V_{K}^{j}$ is a Cauchy sequence in $A=\mathscr{L}(C(I))$, and $u=\sum_{n=0}^{\infty}V_{K}^{n}u_{0}\in C(I)$ satisfies $u=(e-V_{K})^{-1}u_{0}$.

###### Proof.

$V_{K}\in C(I)$ is quasinilpotent so applying (1) with $\lambda=1$,

 $(e-V_{K})^{-1}=\lim_{n\to\infty}\sum_{j=0}^{n}V_{K}^{j}\in A.$

Then

 $(e-V_{K})^{-1}u_{0}=\left(\lim_{n\to\infty}\sum_{j=0}^{n}V_{K}^{j}\right)u_{0}% =\lim_{n\to\infty}(V_{K}^{j}u_{0})=\sum_{n=0}^{\infty}V_{K}^{n}u_{0}.$

Hence $u=(1-V_{K})^{-1}u_{0}$, and so $(1-V_{K})u=u_{0}$, i.e. $u=u_{0}+V_{K}u$, i.e. for $x\in I$,

 $\displaystyle u(x)$ $\displaystyle=u_{0}(x)+\int_{0}^{x}K(x,y)u(y)dy.$

###### Theorem 8.

Let $Q\in C(I)$ and let $K(x,y)=(x-y)Q(y)$, $K\in C(I\times I)$. Let $u_{0}(x)=x$, $u_{0}\in C(I)$. Then $\sum_{j=0}^{n}V_{K}^{j}$ is a Cauchy sequence in $A=\mathscr{L}(C(I))$, and $u=\sum_{n=0}^{\infty}V_{K}^{n}u_{0}\in C(I)$ satisfies $u\in C^{2}(I)$,

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1.$
###### Proof.

By Lemma 7, $u=(e-V_{K})^{-1}u_{0}$, i.e. $(e-V_{K})u=u_{0}$, i.e. $u-V_{K}u=u_{0}$, i.e. for $x\in I$,

 $u(x)=x+V_{K}u(x)=x+\int_{0}^{x}K(x,y)u(y)dy=x+\int_{0}^{x}(x-y)Q(y)u(y)dy.$

Lemma 6 then tells us that $u\in C^{2}(I)$ and

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1.$

## 4 Gronwall’s inequality

Let $f\in L^{1}(I)$. We say that $x\in I$ is a Lebesgue point of $f$ if

 $\frac{1}{r}\int_{x}^{x+r}|f(y)-f(x)|dy\to 0,\qquad r\to 0,$

which implies

 $\frac{1}{r}\int_{x}^{x+r}f(y)dy\to f(x),\qquad r\to 0.$

The Lebesgue differentiation theorem77 7 Walter Rudin, Real and Complex Analysis, third ed., p. 138, Theorem 7.7 states that for almost all $x\in I$, $x$ is a Lebesgue point of $f$. Let

 $F(x)=\int_{0}^{x}f(y)dy,\qquad x\in I,$

so

 $F(x+r)-F(x)=\int_{x}^{x+r}f(y)dy.$

If $x$ is a Lebesgue point of $f$ then

 $\frac{F(x+r)-F(x)}{r}=\frac{1}{r}\int_{x}^{x+r}f(y)dy\to f(x),$

which means that if $x$ is a Lebesgue point of $f$ then

 $F^{\prime}(x)=f(x).$

We now prove Gronwall’s inequality.88 8 Anton Zettl, Sturm-Liouville Theory, p. 8, Theorem 1.4.1.

###### Theorem 9 (Gronwall’s inequality).

Let $g\in L^{1}(I)$, $g\geq 0$ almost everywhere and let $f:I\to\mathbb{R}$ be continuous. If $y:I\to\mathbb{R}$ is continuous and

 $y(t)\leq f(t)+\int_{0}^{t}g(s)y(s)ds,\qquad t\in I,$

then

 $y(t)\leq f(t)+\int_{0}^{t}f(s)g(s)\exp\left(\int_{s}^{t}g(u)du\right)ds,\qquad t% \in I.$

If $f$ is increasing then

 $y(t)\leq f(t)\exp\left(\int_{0}^{t}g(s)ds\right),\qquad t\in I.$
###### Proof.

Let $z(t)=g(t)y(t)$ and

 $Z(t)=\int_{0}^{t}z(s)ds,\qquad t\in I.$

By hypothesis, $g\geq 0$ almost everywhere, and by the Lebesgue differentiation theorem, $Z^{\prime}(t)=z(t)$ for almost all $t\in I$. Therefore for almost all $t\in I$,

 $Z^{\prime}(t)=z(t)=g(t)y(t)\leq g(t)\left(f(t)+\int_{0}^{t}g(s)y(s)ds\right)=g% (t)f(t)+g(t)Z(t).$

That is, there is a Borel set $E\subset I$, $\mu(E)=1$, such that for $t\in I$, $Z$ is differentiable at $t$ and

 $Z^{\prime}(t)-g(t)Z(t)\leq g(t)f(t).$

For $s\in E$, using the product rule,

 $\displaystyle\bigg{[}\exp\left(-\int_{0}^{s}g(u)du\right)Z(s)\bigg{]}^{\prime}$ $\displaystyle=\exp\left(-\int_{0}^{s}g(u)du\right)\bigg{[}Z^{\prime}(s)-g(t)Z(% s)\bigg{]}.$

For $t\in I$, as $\mu(E)=1$,

 $\begin{split}&\displaystyle\int_{0}^{t}\bigg{[}\exp\left(-\int_{0}^{s}g(u)du% \right)Z(s)\bigg{]}^{\prime}ds\\ \displaystyle=&\displaystyle\int_{[0,t]\cap E}\bigg{[}\exp\left(-\int_{0}^{s}g% (u)du\right)Z(s)\bigg{]}^{\prime}ds\\ \displaystyle=&\displaystyle\int_{[0,t]\cap E}\exp\left(-\int_{0}^{s}g(u)du% \right)\bigg{[}Z^{\prime}(s)-g(s)Z(s)\bigg{]}ds\\ \displaystyle\leq&\displaystyle\int_{[0,t]\cap E}\exp\left(-\int_{0}^{s}g(u)du% \right)g(s)f(s)ds\\ \displaystyle=&\displaystyle\int_{0}^{t}g(s)f(s)\exp\left(-\int_{0}^{s}g(u)du% \right)ds.\end{split}$

But

 $\displaystyle\int_{0}^{t}\bigg{[}\exp\left(-\int_{0}^{s}g(u)du\right)Z(s)\bigg% {]}^{\prime}ds$ $\displaystyle=\bigg{[}\exp\left(-\int_{0}^{s}g(u)du\right)Z(s)\bigg{]}\bigg{|}% _{0}^{t}$ $\displaystyle=\exp\left(-\int_{0}^{t}g(u)du\right)Z(t).$

So

 $\exp\left(-\int_{0}^{t}g(u)du\right)Z(t)\leq\int_{0}^{t}g(s)f(s)\exp\left(-% \int_{0}^{s}g(u)du\right)ds.$

Therefore,

 $\displaystyle y(t)$ $\displaystyle\leq f(t)+\int_{0}^{t}g(s)y(s)ds$ $\displaystyle=f(t)+Z(t)$ $\displaystyle\leq f(t)+\exp\left(\int_{0}^{t}g(u)du\right)\int_{0}^{t}g(s)f(s)% \exp\left(-\int_{0}^{s}g(u)du\right)ds$ $\displaystyle=f(t)+\int_{0}^{t}g(s)f(s)\exp\left(\int_{0}^{t}g(u)du-\int_{0}^{% s}g(u)du\right)ds$ $\displaystyle=f(t)+\int_{0}^{t}g(s)f(s)\exp\left(\int_{s}^{t}g(u)du\right)ds.$

Suppose that $f$ is increasing. Let

 $G(s)=\int_{0}^{s}g(u)du,\qquad s\in I.$

For $t\in I$,

 $\displaystyle y(t)$ $\displaystyle\leq f(t)+\int_{0}^{t}g(s)f(s)\exp\left(\int_{s}^{t}g(u)du\right)ds$ $\displaystyle\leq f(t)+\int_{0}^{t}g(s)f(t)\exp\left(\int_{s}^{t}g(u)du\right)ds$ $\displaystyle=f(t)\bigg{[}1+\int_{0}^{t}g(s)\exp\left(\int_{s}^{t}g(u)du\right% )ds\bigg{]}$ $\displaystyle=f(t)\bigg{[}1+\int_{0}^{t}g(s)e^{G(t)-G(s)}ds\bigg{]}$ $\displaystyle=f(t)\bigg{[}1+e^{G(t)}\int_{0}^{t}g(s)e^{-G(s)}ds\bigg{]}.$

Let $H(s)=e^{-G(s)}$, with which

 $y(t)\leq f(t)\bigg{[}1+\frac{1}{H(t)}\int_{0}^{t}g(s)H(s)ds\bigg{]}.$

If $s$ is a Lebesgue point of $g$ then

 $H^{\prime}(s)=-G^{\prime}(s)e^{-G(s)}=-g(s)H(s).$

Hence

 $\displaystyle y(t)$ $\displaystyle\leq f(t)\bigg{[}1-\frac{1}{H(t)}\int_{0}^{t}H^{\prime}(s)ds\bigg% {]}$ $\displaystyle=f(t)\bigg{[}1-\frac{1}{H(t)}\bigg{[}H(t)-H(0)\bigg{]}\bigg{]}$ $\displaystyle=f(t)\bigg{[}1-1+\frac{H(0)}{H(t)}\bigg{]}$ $\displaystyle=f(t)e^{G(t)}$ $\displaystyle=f(t)\exp\left(\int_{0}^{t}g(u)du\right).$

Let $K(x,y)=(x-y)Q(y)$. Let $u=\sum_{n=0}^{\infty}V_{K}^{n}u_{0}\in C(I)$. Lemma 7 tells us that $u=(e-V_{K})^{-1}u_{0}$, i.e. $(e-V_{K})u=u_{0}$, i.e. $u=u_{0}+V_{K}u$, i.e. for $x\in I$,

 $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy.$

Then

 $|u(x)|\leq x+\int_{0}^{x}|x-y||Q(y)||u(y)|dy\leq x+\int_{0}^{x}|Q(y)||u(y)|dy.$

Applying Gronwall’s inequality we get

 $|u(x)|\leq x\exp\left(\int_{0}^{x}|Q(y)|dy\right),\qquad x\in I.$ (2)

## 5 The spectral theorem for positive compact operators

The following is the spectral theorem for positive compact operators.99 9 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 102, Theorem 3.2.1.

###### Theorem 10 (Spectral theorem for positive compact operators).

Let $H$ be a separable complex Hilbert space and let $T\in\mathscr{L}(H)$ be positive and compact. There are countable sets $\Phi,\Psi\subset H$ and $\lambda_{\phi}>0$ for $\phi\in\Phi$ such that (i) $\Phi\cup\Psi$ is an orthonormal basis for $H$, (ii) $T\phi=\lambda_{\phi}\phi$ for $\phi\in\Phi$, (iii) $T\psi=0$ for $\psi\in\Psi$, (iv) if $\Phi$ is infinite then $0$ is a limit point of $\Lambda$ and is the only limit point of $\Lambda$.

Suppose that $H$ is infinite dimensional and that $T$ is a positive compact operator with $\ker(T)=0$. The spectral theorem for positive compact operators then says that there is a a countable set $\Phi\subset H$ and $\lambda_{\phi}>0$ for $\phi\in\Phi$ such that $\Phi$ is an orthonormal basis for $H$, $T\phi=\lambda_{\phi}\phi$ for $\phi\in\Phi$, and the unique limit point of $\{\lambda_{\phi}:\phi\in\Phi\}$ is $0$. Let $\Phi=\{\phi_{n}:n\geq 1\}$, $\phi_{n}\neq\phi_{m}$ for $n\geq m$, such that $n\geq m$ implies $\lambda_{\phi_{n}}\leq\lambda_{\phi_{m}}$. Let $\lambda_{n}=\lambda_{\phi_{n}}$. Then $\lambda_{n}\downarrow 0$. Summarizing, there is an orthonormal basis $\{\phi_{n}:n\geq 1\}$ for $H$ and $\lambda_{n}>0$ such that $T\phi_{n}=\lambda_{n}\phi_{n}$ for $n\geq 1$ and $\lambda_{n}\downarrow 0$.

## 6 $Q>0$, Green’s function for $L_{Q}$

Suppose $Q\in C(I)$ with $Q(x)>0$ for $0. Let $K(x,y)=(x-y)Q(y)$, $K\in C(I\times I)$, and $u_{0}(x)=x$, $u_{0}\in C(I)$. Let

 $u=\sum_{n=0}^{\infty}V_{K}^{n}u_{0}\in C(I).$

By Theorem 8, $u\in C^{2}(I)$ and

 $L_{Q}u=0,\qquad u(0)=0,\quad u^{\prime}(0)=1.$

If $f\in C(I)$ and $f(x)>0$ for $0 then

 $V_{K}f(x)=\int_{0}^{x}(x-y)Q(y)f(y)dy>0.$

By induction, for $0 and for $n\geq 1$ we have $V_{K}^{n}f(x)>0$. Hence for $0,

 $u(x)=\sum_{n=0}^{\infty}(V_{K}^{n}u_{0})(x)>0.$

For $x\in I$,

 $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy=x+x\int_{0}^{x}Q(y)u(y)dy-\int_{0}^{x}yQ(y)% u(y)dy.$

Using the fundamental theorem of calculus,

 $u^{\prime}(x)=1+\int_{0}^{x}Q(y)u(y)dy.$

Then because $Q(y)>0$ for $0 and $u(y)>0$ for $0,

 $u^{\prime}(x)>1,\qquad 0

Using $u(x)=x+\int_{0}^{x}(x-y)Q(y)u(y)dy$ and $Q>0$ we get

 $u(x)>x,\qquad 0

Let $u_{1}(x)=u(x)$ and $u_{2}(x)=u(1-x)$. Then

 $L_{Q}u_{1}=0,\qquad u_{1}(0)=0,\qquad u_{1}^{\prime}(0)=1$

and

 $L_{Q}u_{2}=0,\qquad u_{2}(1)=0,\qquad u_{2}^{\prime}(1)=-1.$

A fortiori,

 $u_{1}(x)>0,\qquad u_{1}^{\prime}(x)>0,\qquad 0

and as $u_{2}^{\prime}(x)=-u^{\prime}(1-x)$,

 $u_{2}(x)>0,\qquad u_{2}^{\prime}(x)<0,\qquad 0

For $0 let

 $W(x)=u_{1}^{\prime}(x)u_{2}(x)-u_{1}(x)u_{2}^{\prime}(x).$

$u_{1}^{\prime}>0,u_{2}>0$ so $u_{1}^{\prime}u_{2}>0$. $u_{1}>0,u_{2}^{\prime}<0$ so $-u_{1}u_{2}^{\prime}>0$, hence $W>0$.

 $\displaystyle W^{\prime}$ $\displaystyle=(u_{1}^{\prime}u_{2}-u_{1}u_{2}^{\prime})^{\prime}$ $\displaystyle=u_{1}^{\prime\prime}u_{2}+u_{1}^{\prime}u_{2}^{\prime}-u_{1}^{% \prime}u_{2}^{\prime}-u_{1}u_{2}^{\prime\prime}$ $\displaystyle=u_{1}^{\prime\prime}u_{2}-u_{1}u_{2}^{\prime\prime}$ $\displaystyle=(Qu_{1})u_{2}-u_{1}(Qu_{2})$ $\displaystyle=0.$

Therefore there is some $W_{0}>0$ such that $W(x)=W_{0}$ for all $0.

Define

 $G(x,y)=\frac{u_{1}(x\wedge y)u_{2}(x\vee y)}{W_{0}},\qquad(x,y)\in I\times I.$

$x\wedge y=\min(x,y)$, $x\vee y=\max(x,y)$. Because $(x,y)\mapsto x\wedge y$ and $(x,y)\mapsto x\vee y$ are each continuous $I\times I\to I$, it follows that $G\in C(I\times I)$. $G(x,y)=G(y,x)$.

$G$ is the Green’s function for $L_{Q}$. Let $(x,y)\in I\times I$. If $x>y$ then

 $G^{y}(x)=\frac{u_{1}(y)u_{2}(x)}{W_{0}}$

and so

 $L_{Q}G^{y}(x)=\frac{u_{1}(y)}{W_{0}}L_{Q}u_{2}(x)=0.$

If $x then

 $G^{y}(x)=\frac{u_{1}(x)u_{2}(y)}{W_{0}}$

and so

 $L_{Q}G^{y}(x)=\frac{u_{2}(y)}{W_{0}}L_{Q}u_{1}(x)=0.$

## 7 $Q>0$, $L^{2}(I)$

$L^{2}(I)$ is a separable complex Hilbert space with the inner product

 $\left\langle f,g\right\rangle=\int_{I}f\overline{g}d\mu,\qquad f,g\in L^{2}(I).$

Define $T_{Q}:L^{2}(I)\to L^{2}(I)$ by

 $(T_{Q}g)(x)=\int_{I}G(x,y)g(y)dy.$

$T_{Q}:L^{2}(I)\to L^{2}(I)$ is a Hilbert-Schmidt operator.1010 10 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 96, Theorem 3.1.16.

It is immediate that $G(y,x)=G(x,y)$ and $\overline{G}=G$. Then by Fubini’s theorem, for $f,g\in L^{2}(I)$,

 $\displaystyle\left\langle T_{Q}g,f\right\rangle$ $\displaystyle=\int_{I}(T_{Q}g)(x)\overline{f(x)}dx$ $\displaystyle=\int_{I}\left(\int_{I}G(x,y)g(y)dy\right)\overline{f(x)}dx$ $\displaystyle=\int_{I}g(y)\overline{\left(\int_{I}G(y,x)f(x)dx\right)}dy$ $\displaystyle=\int_{I}g(y)\overline{(T_{Q}f)(y)}dy$ $\displaystyle=\left\langle g,T_{Q}f\right\rangle.$

Therefore $T_{Q}:L^{2}(I)\to L^{2}(I)$ is self-adjoint.

We now establish properties of $T_{Q}$.1111 11 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 106, Proposition 3.2.8. Let

 $N^{k}(I)=\{f\in C^{k}(I):f(0)=0,f(1)=0\}.$
###### Lemma 11.

Let $Q\in C(I)$, $Q(x)>0$ for $0. Let $g\in L^{2}(I)$ and let $f=T_{Q}g$,

 $f(x)=(T_{Q}g)(x)=\int_{I}G(x,y)g(y)dy=\int_{I}G_{x}gd\mu.$

Then $f\in N^{0}(I)$.

If $g\in C(I)$ then $f\in C^{2}(I)$ and

 $L_{Q}f=g.$
###### Proof.

For $x\in I$,

 $\displaystyle f(x)$ $\displaystyle=\int_{0}^{x}\frac{u_{1}(x\wedge y)u_{2}(x\vee y)}{W_{0}}g(y)dy+% \int_{x}^{1}\frac{u_{1}(x\wedge y)u_{2}(x\vee y)}{W_{0}}g(y)dy$ $\displaystyle=\int_{0}^{x}\frac{u_{1}(y)u_{2}(x)}{W_{0}}g(y)dy+\int_{x}^{1}% \frac{u_{1}(x)u_{2}(y)}{W_{0}}g(y)dy$ $\displaystyle=u_{2}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+u_{1}(x)\int_{x% }^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy.$

It follows that $f\in C(I)$.

Suppose $g\in C(I)$. Then by the fundamental theorem of calculus,

 $\displaystyle f^{\prime}(x)$ $\displaystyle=u_{2}^{\prime}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+u_{2}(% x)\frac{u_{1}(x)g(x)}{W_{0}}$ $\displaystyle+u_{1}^{\prime}(x)\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy-u_{1}(% x)\frac{u_{2}(x)g(x)}{W_{0}}$ $\displaystyle=u_{2}^{\prime}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+u_{1}^% {\prime}(x)\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy.$

Because $u_{1}^{\prime},u_{2}^{\prime}\in C(I)$ it follows that $f^{\prime}\in C(I)$, i.e. $f\in C^{1}(I)$. Then

 $\displaystyle f^{\prime\prime}(x)$ $\displaystyle=u_{2}^{\prime\prime}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+% u_{2}^{\prime}(x)\frac{u_{1}(x)g(x)}{W_{0}}$ $\displaystyle+u_{1}^{\prime\prime}(x)\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy-% u_{1}^{\prime}(x)\frac{u_{2}(x)g(x)}{W_{0}}$ $\displaystyle=u_{2}^{\prime\prime}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+% u_{1}^{\prime\prime}(x)\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy-\frac{W(x)g(x)% }{W_{0}}$ $\displaystyle=u_{2}^{\prime\prime}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+% u_{1}^{\prime\prime}(x)\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy-g(x).$

Because $g\in C(I)$ it follows that $f^{\prime\prime}\in C(I)$, i.e. $f\in C^{2}(I)$. Furthermore, because $u_{1}^{\prime\prime}=Qu_{1}$ and $u_{2}^{\prime\prime}=Qu_{2}$,

 $\displaystyle f^{\prime\prime}(x)$ $\displaystyle=Q(x)u_{2}(x)\int_{0}^{x}\frac{u_{1}(y)g(y)}{W_{0}}dy+Q(x)u_{1}(x% )\int_{x}^{1}\frac{u_{2}(y)g(y)}{W_{0}}dy-g(x)$ $\displaystyle=Q(x)f(x)-g(x).$

We now establish more facts about $T_{Q}$.1212 12 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 107, Proposition 3.2.9.

###### Lemma 12.

Let $Q\in C(I)$, $Q(x)>0$ for $0.

1. 1.

If $f_{1},f_{2}\in N^{2}(I)$ then

 $\int_{I}f_{1}L_{Q}f_{2}dx=\int_{I}(f_{1}^{\prime}f_{2}^{\prime}+Qf_{1}f_{2})dx.$
2. 2.

If $f\in N^{2}(I)$ and $L_{Q}f=0$, then $f=0$.

3. 3.

If $f\in N^{2}(I)$ then $f=T_{Q}L_{Q}f$.

4. 4.

$T_{Q}\geq 0$.

5. 5.

$\ker T_{Q}=0$.

###### Proof.

First, doing integration by parts,

 $\displaystyle\int_{I}f_{1}(-f_{2}^{\prime\prime}+Qf_{2})dx$ $\displaystyle=-\int_{\partial I}f_{1}f_{2}^{\prime}+\int_{I}f_{1}^{\prime}f_{2% }^{\prime}dx+\int_{I}Qf_{1}f_{2}dx$ $\displaystyle=\int_{I}f_{1}^{\prime}f_{2}^{\prime}dx+\int_{I}Qf_{1}f_{2}dx$ $\displaystyle=\int_{I}(f_{1}^{\prime}f_{2}^{\prime}+Qf_{1}f_{2})dx.$

Second, using the above with $f_{1}=f$ and $f_{2}=f$, with $f\in C^{2}(I)$ real-valued,

 $\int_{I}f(-f^{\prime\prime}+Qf)dx=\int_{I}(|f^{\prime}|^{2}+Q|f|^{2})dx.$

Using $-f^{\prime\prime}+Qf=0$,

 $\int_{I}(|f^{\prime}|^{2}+Q|f|^{2})dx=0.$

Because $Q(x)>0$ for $0, it follows that $|f|=0$ almost everywhere. But $f$ is continuous so $f=0$. For $f=f_{1}+if_{2}$, if $-f^{\prime\prime}+Qf=0$ and $f(0)=0,f(1)=0$ then as $Q$ is real-valued, we get $f_{1}=0$ and $f_{2}=0$ hence $f=0$.

Third, say $f\in C^{2}(I)$ is real-valued, $f(0)=0$, $f(1)=0$, and $g=L_{Q}f=-f^{\prime\prime}+Qf\in C(I)$. Let $h=T_{Q}g$. By Lemma 11, $h\in C^{2}(I)$ and

 $-h^{\prime\prime}+Qh=g,\qquad h(0)=0,\qquad h(1)=0.$

Let $F=f-h$. Then using $-f^{\prime\prime}+Qf=g$ we get

 $F^{\prime\prime}=f^{\prime\prime}-h^{\prime\prime}=(Qf-g)-(Qh-g)=Q(f-h)=QF.$

Furthermore,

 $F(0)=f(0)-h(0)=0-0=0,\qquad F(1)=f(1)-h(1)=0-0=0.$

Because $f$ is real-valued so is $g$, and because $g$ is real-valued it follows that $h=T_{Q}g$ is real-valued. Thus $F$ is real-valued and so by the above, $F=0$. That is, $f=h$, i.e. $f=T_{Q}g$. For $f=f_{1}+if_{2}$, if $f(0)=0$, $f(1)=0$ and $g=-f^{\prime\prime}+Qf$, let $g=g_{1}+ig_{2}$. As $Q$ is real-valued we get $g_{1}=-f_{1}^{\prime\prime}+Qf_{1}$ and $g_{2}=-f_{2}^{\prime\prime}+Qf_{2}$. Then $f_{1}=T_{Q}g_{1}$ and $f_{2}=T_{Q}g_{2}$. Thus

 $f=f_{1}+if_{2}=T_{Q}g_{1}+iT_{Q}g_{2}=T_{Q}(g_{1}+ig_{2})=T_{Q}g.$

Fourth, let $g\in C(I)$ and let $f=T_{Q}g$. By Lemma 11, $f\in C^{2}(I)$ and

 $-f^{\prime\prime}+Qf=g,\qquad f(0)=0,\qquad f(1)=0.$

Then using the above,

 $\displaystyle\left\langle g,T_{Q}g\right\rangle$ $\displaystyle=\left\langle-f^{\prime\prime}+Qf,f\right\rangle$ $\displaystyle=\int_{I}(-f^{\prime\prime}+Qf)\overline{f}dx$ $\displaystyle=\int_{I}(\overline{f}^{\prime}f^{\prime}+Q\overline{f}f)dx$ $\displaystyle=\int_{I}(|f^{\prime}|^{2}+Q|f|^{2})dx.$

Because $Q\geq 0$ we have $\left\langle g,T_{Q}g\right\rangle\geq 0$. For $g\in L^{2}(I)$ let $g_{n}\in C(I)$ with $\left\|g_{n}-g\right\|_{L^{2}}\to 0$. Then $\left\langle g_{n},T_{Q}g_{n}\right\rangle\to\left\langle g,T_{Q}g\right\rangle$ as $n\to\infty$, and because $\left\langle g_{n},T_{Q}g_{n}\right\rangle\geq 0$ it follows that $\left\langle g,T_{Q}g\right\rangle\geq 0$. Therefore $T_{Q}\geq 0$, namely $T_{Q}$ is a positive operator.

Let $f\in N^{2}$ and let $g=-f^{\prime\prime}+Qf$. Then $f=T_{Q}g$. This means that $N^{2}\subset\mathrm{Ran}(T_{Q})$. One checks that $N^{2}$ is dense in $L^{2}(I)$, so $\mathrm{Ran}(T_{Q})$ is dense in $L^{2}(I)$. If $f\in\ker(T_{Q})$ and $g\in L^{2}(I)$ then $\left\langle f,T_{Q}^{*}g\right\rangle=\left\langle T_{Q}f,g\right\rangle=0$. Hence $\ker(T_{Q})\perp\mathrm{Ran}(T_{Q}^{*})$. But $T_{Q}$ is self-adjoint which implies that $\ker(T_{Q})\perp\mathrm{Ran}(T_{Q})$. Because $\mathrm{Ran}(T_{Q})$ is dense in $L^{2}(I)$ it follows that $\ker(T_{Q})=0$. ∎

We now prove the Sturm-Liouville theorem.1313 13 Barry Simon, Operator Theory. A Comprehensive Course in Analysis, Part 4, p. 105, Theorem 3.2.7, p. 110, Exercise 7.

###### Theorem 13 (Sturm-Liouville theorem).

Let $Q\in C(I)$, $Q(x)>0$ for $0. There is an orthonormal basis $\{u_{n}:n\geq 1\}\subset N^{2}(I)$ for $L^{2}(I)$ and $\lambda_{n}>0$, $\lambda_{m}<\lambda_{n}$ for $m and $\lambda_{n}\to\infty$, such that

 $L_{Q}u_{n}=\lambda_{n}u_{n},\qquad n\geq 1.$
###### Proof.

We have established that $T_{Q}$ is a positive compact operator with $\ker T_{Q}=0$. The spectral theorem for positive compact operators then tells us that there is an orthonormal basis $\{\phi_{n}:n\geq 1\}$ for $L^{2}(I)$ and $\gamma_{n}>0$ such that $T_{Q}\phi_{n}=\gamma_{n}\phi_{n}$ for $n\geq 1$ and $\gamma_{n}\downarrow 0$. By Lemma 11, $T_{Q}\phi_{n}\in N^{0}(I)$. Let

 $u_{n}=\frac{1}{\gamma_{n}}T_{Q}\phi_{n}\in N^{0}(I).$

Because $T_{Q}\phi_{n}=\gamma_{n}\phi_{n}$ we have $u_{n}=\phi_{n}$ in $L^{2}(I)$ and so

 $u_{n}=\frac{1}{\gamma_{n}}T_{Q}u_{n}.$

Let $v_{n}=T_{Q}u_{n}$. Because $u_{n}\in C(I)$, Lemma 11 tells us that $v_{n}\in N^{2}(I)$ and $L_{Q}v_{n}=u_{n}$. But $u_{n}=\frac{1}{\gamma_{n}}v_{n}$ so $u_{n}\in N^{2}(I)$ and

 $L_{Q}u_{n}=\frac{1}{\gamma_{n}}L_{Q}v_{n}=\frac{1}{\gamma_{n}}u_{n}.$

Let $\lambda_{n}=\frac{1}{\gamma_{n}}$. Then $\lambda_{n}>0$, $\lambda_{m}\leq\lambda_{n}$ for $m\leq n$, $\lambda_{n}\to\infty$, and

 $L_{Q}u_{n}=\lambda_{n}u_{n},\qquad n\geq 1.$

To prove the claim it remains to show that the sequence $\lambda_{n}$ is strictly increasing.

Let $\lambda>0$ and suppose that $f,g\in N^{2}(I)$ satisfy

 $L_{Q}f=\lambda f,\qquad L_{Q}g=\lambda g.$

Let $W(x)=f(x)g^{\prime}(x)-g(x)f^{\prime}(x)$, the Wronskian of $f$ and $g$. Either $W(x)=0$ for all $x\in I$ or $W(x)\neq 0$ for all $x\in I$. Using $f(0)=0$ and $g(0)=0$ we get $W(0)=0$. Therefore $W(x)=0$ for all $x\in I$ and $W=0$ implies that $f,g$ are linearly dependent.

Suppose by contradiction that $\lambda_{n}=\lambda_{m}$ for some $n\neq m$. Applying the above with $\lambda=\lambda_{n}=\lambda_{m}$, $f=u_{n},g=u_{m}$ we get that $u_{n},u_{m}$ are linearly dependent, contradicting that $\{u_{n}:n\geq 1\}$ is an orthonormal set. Therefore $m\neq n$ implies that $\lambda_{m}\neq\lambda_{n}$. ∎

## 8 Other results in Sturm-Liouville theory

1414 14 B. M. Levitan and I. S. Sargsjan, Spectral Theory: Selfadjoint Ordinary Differential Operators, p. 11.