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

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

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

The spectrum of $x$ is

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

The spectral radius of $x$ is

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

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

$$r(x)=\underset{n\to \mathrm{\infty }}{lim}{\parallel x^n\parallel }^{1/n},$$

the spectral radius formula. If $r(x)=0$ we say that $x$ is quasinilpotent. $x\in A$ is quasinilpotent if and only if $\sigma (x)=\{0\}$.

Lemma 1.

If $x\mathrm{\in }A$ is quasinilpotent and $\mathrm{|}\lambda \mathrm{|}\mathrm{>}\mathrm{0}$, then $S_n\mathrm{=}{\mathrm{\sum }}_{j\mathrm{=}\mathrm{0}}^n{\lambda }^j\mathit{}{x}^j\mathrm{\in }A$ is a Cauchy sequence, and

$$(e-\lambda x)\sum _{n=0}^{\mathrm{\infty }}{\lambda }^n{x}^n=e.$$

Proof.

Let . There is some $n_{ϵ}$ such that ${\parallel x^n\parallel }^{1/n}\le ϵ$ for $n\ge n_{ϵ}$. For $n>m\ge n_{ϵ}$,

$$\parallel S_n-S_m\parallel \le \sum _{j=m+1}^n{|\lambda |}^j\parallel x^j\parallel \le \sum _{j=m+1}{|\lambda |}^j{ϵ}^j,$$

and because , it follows that $S_n\in A$ is a Cauchy sequence and so converges to some $S\in A$, $S={\sum }_{n=0}^{\mathrm{\infty }}{\lambda }^k{x}^k$. Now,

$(e-\lambda x)S$ $=(e-\lambda x)S_n+(e-\lambda x)(S-S_n)$

$=S_n-\lambda x{S}_n+(e-\lambda x)(S-S_n)$

$=S_n-{\displaystyle \sum _{j=1}^{n+1}}{\lambda }^j{x}^j+(e-\lambda x)(S-S_n)$

$=e-{\lambda }^{n+1}{x}^{n+1}+(e-\lambda x)(S-S_n).$

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

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

$$R_x(\lambda )={(x-\lambda e)}^{-1}.$$

Lemma 2.

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

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

and if $\mathrm{|}\lambda \mathrm{|}\mathrm{>}\mathrm{0}$ then

$$R_x(\lambda )=-{\lambda }^{-1}{(e-{\lambda }^{-1}x)}^{-1}=-{\lambda }^{-1}\sum _{n=0}^{\mathrm{\infty }}{\lambda }^{-n}{x}^n.$$

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

$${\parallel f\parallel }_{\mathrm{\infty }}=\underset{x\in I}{sup}|f(x)|,f\in C(I).$$

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

$${\parallel f\parallel }_{L^1}={\int }_I|f(x)|𝑑x,f\in L^1(I).$$

For $f:I\to ℂ$, let

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

Let $\mathrm{Lip}(I)$ be the set of those $f:I\to ℂ$ with . It is a fact that $\mathrm{Lip}(I)$ is a Banach space with the norm ${\parallel f\parallel }_{\mathrm{Lip}}={\parallel f\parallel }_{\mathrm{\infty }}+{|f|}_{\mathrm{Lip}}$.

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

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

$$\parallel T\parallel =\underset{f\in C(I),{\parallel f\parallel }_{\mathrm{\infty }}\le 1}{sup}{\parallel Tf\parallel }_{\mathrm{\infty }},T\in A.$$

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

$$K_x(y)=K(x,y),K^y(x)=K(x,y).$$

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

$$V_{K}f(x)={\int }_0^{x}K(x,y)f(y)𝑑y,x\in I.$$

Lemma 3.

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

Proof.

For $x_1,x_2\in I$, $x_1>x_2$,

$V_{K}f(x_1)-V_{K}f(x_2)$ $={\displaystyle {\int }_0^{x_1}}K(x_1,y)f(y)𝑑y-{\displaystyle {\int }_0^{x_1}}K(x_2,y)f(y)𝑑y$

$+{\displaystyle {\int }_0^{x_1}}K(x_2,y)f(y)𝑑y-{\displaystyle {\int }_0^{x_2}}K(x_2,y)f(y)𝑑y$

$={\displaystyle {\int }_0^{x_1}}\left[K(x_1,y)-K(x_2,y)\right]f(y)𝑑y+{\displaystyle {\int }_{x_2}^{x_1}}K(x_2,y)f(y)𝑑y.$

Let $ϵ>0$. Because $K:I\times I\to ℂ$ is uniformly continuous, there is some ${\delta }_1>0$ such that $|(x_1,y_1)-(x_2,y_2)|\le {\delta }_1$ implies $|K(x_1,y_1)-K(x_2,y_2)|\le ϵ$. By the absolute continuity of the Lebesgue integral, there is some ${\delta }_2>0$ such that $\mu (E)\le {\delta }_2$ implies ${\int }_E|f|𝑑\mu \le ϵ$. Therefore if then

$|V_{K}f(x_1)-V_{K}f(x_2)|$ $\le {\displaystyle {\int }_0^{x_1}}ϵ|f(y)|𝑑y+{\parallel K\parallel }_{\mathrm{\infty }}{\displaystyle {\int }_{x_2}^{x_1}}|f(y)|𝑑y$

$\le ϵ{\parallel f\parallel }_{L^1}+{\parallel K\parallel }_{\mathrm{\infty }}ϵ.$

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

${\parallel V_{K}f\parallel }_{\mathrm{\infty }}\le {\parallel K\parallel }_{\mathrm{\infty }}{\parallel f\parallel }_{\mathrm{\infty }}$ so $\parallel V_K\parallel \le {\parallel K\parallel }_{\mathrm{\infty }}$, 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 $x\in I$,

$$V_K^{2}f(x)={\int }_0^{x}K(x,y_1)V_{K}f(y_1)𝑑y_1={\int }_0^{x}K(x,y_1)\left({\int }_0^{y_1}K(y_1,y_2)f(y_2)𝑑y_2\right)𝑑y_1.$$

$V_K^{3}f(x)$ $=V_K^2{V}_{K}f(x)$

$={\displaystyle {\int }_0^x}K(x,y_1){\displaystyle {\int }_0^{y_1}}K(y_1,y_2)V_{K}f(y_2)𝑑y_2𝑑y_1$

$={\displaystyle {\int }_0^x}K(x,y_1){\displaystyle {\int }_0^{y_1}}K(y_1,y_2){\displaystyle {\int }_0^{y_2}}K(y_2,y_3)f(y_3)𝑑y_3𝑑y_2𝑑y_1.$

For $n\ge 2$,

$$V_K^{n}f(x)={\int }_{y_1=0}^x{\int }_{y_2=0}^{y_1}\mathrm{\cdots }{\int }_{y_n=0}^{y_{n-1}}K(x,y_1)K(y_1,y_2)\mathrm{\cdots }K(y_{n-1},y_n)f(y_n)𝑑y_n\mathrm{\cdots }𝑑y_1.$$

We prove that $V_K$ is quasinilpotent.

Theorem 4.

If $K\mathrm{\in }C\mathit{}\mathrm{(}I\mathrm{\times }I\mathrm{)}$ then

$$\parallel V_K^n\parallel \le \frac{{\parallel K\parallel }_{\mathrm{\infty }}^n}{n!},$$

and thus $V_K\mathrm{\in }A\mathrm{=}\mathrm{L}\mathit{}\mathrm{(}C\mathit{}\mathrm{(}I\mathrm{)}\mathrm{)}$ is quasinilpotent.

Proof.

Let

${\mathrm{\Phi }}_n(x)$ $={\displaystyle {\int }_0^x}{\displaystyle {\int }_0^{y_1}}\mathrm{\cdots }{\displaystyle {\int }_0^{y_{n-1}}}𝑑y_n\mathrm{\cdots }𝑑y_1$

$={\displaystyle {\int }_0^x}{\displaystyle {\int }_0^{y_1}}\mathrm{\cdots }{\displaystyle {\int }_0^{y_{n-2}}}{y}_{n-1}𝑑y_{n-1}\mathrm{\cdots }𝑑y_1$

$={\displaystyle {\int }_0^x}{\displaystyle {\int }_0^{y_1}}\mathrm{\cdots }{\displaystyle {\int }_0^{y_{n-3}}}{\displaystyle \frac{y_{n-2}^2}{2}}𝑑y_{n-2}\mathrm{\cdots }𝑑y_1$

$={\displaystyle {\int }_0^x}{\displaystyle \frac{y_1^{n-1}}{(n-1)!}}𝑑y_1$

$={\displaystyle \frac{x^n}{n!}}.$

For $x\in I$,

$|V_K^{n}f(x)|$ $\le {\parallel K\parallel }_{\mathrm{\infty }}^n{\parallel f\parallel }_{\mathrm{\infty }}{\displaystyle {\int }_0^x}{\displaystyle {\int }_0^{y_1}}\mathrm{\cdots }{\displaystyle {\int }_0^{y_{n-1}}}𝑑y_n\mathrm{\cdots }𝑑y_1$

$={\parallel K\parallel }_{\mathrm{\infty }}^n{\parallel f\parallel }_{\mathrm{\infty }}{\mathrm{\Phi }}_n(x)$

$={\parallel K\parallel }_{\mathrm{\infty }}^n{\parallel f\parallel }_{\mathrm{\infty }}{\displaystyle \frac{x^n}{n!}}.$

Hence

$$\parallel V_K^n\parallel \le \frac{{\parallel K\parallel }_{\mathrm{\infty }}^n}{n!}.$$

Then

$${\parallel V_K^n\parallel }^{1/n}\le \frac{{\parallel K\parallel }_{\mathrm{\infty }}}{{(n!)}^{1/n}}.$$

Using ${(n!)}^{1/n}\to \mathrm{\infty }$ we get ${\parallel V_K^n\parallel }^{1/n}\to 0$. 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 $\lambda \in ℂ$,

$${(e-\lambda V_K)}^{-1}=\sum _{n=0}^{\mathrm{\infty }}{\lambda }^n{V}_K^n\in A.$$ (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 $u\mathrm{\in }{C}^{\mathrm{2}}\mathit{}\mathrm{(}I\mathrm{)}$ and

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

then

$$u(x)=x+{\int }_0^x(x-y)Q(y)u(y)𝑑y,x\in I.$$

Proof.

For $y\in I$, by the fundamental theorem of calculus 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

$u(x)$ $={\displaystyle {\int }_0^x}{u}^{\prime }(y)𝑑y$

$={\displaystyle {\int }_0^x}\left(1+{\displaystyle {\int }_0^y}Q(t)u(t)𝑑t\right)𝑑y$

$=x+{\displaystyle {\int }_0^x}\left({\displaystyle {\int }_0^y}Q(t)u(t)𝑑t\right)𝑑y.$

Applying Fubini’s theorem,

$u(x)$ $=x+{\displaystyle {\int }_0^x}Q(t)u(t)\left({\displaystyle {\int }_t^x}𝑑y\right)𝑑t$

$=x+{\displaystyle {\int }_0^x}Q(t)u(t)(x-t)𝑑t.$

∎

Lemma 6.

If $u\mathrm{\in }C\mathit{}\mathrm{(}I\mathrm{)}$ and

$$u(x)=x+{\int }_0^x(x-y)Q(y)u(y)𝑑y,x\in I,$$

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

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

Proof.

$$u(x)=x+{\int }_0^x(x-y)Q(y)u(y)𝑑y,x\in I,$$

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,

$u^{\prime }(x)$ $=1+{\displaystyle {\int }_0^x}Q(y)u(y)𝑑y+xQ(x)u(x)-xQ(x)u(x)=1+{\displaystyle {\int }_0^x}Q(y)u(y)𝑑y$

hence

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

and so

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

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

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

∎

Lemma 7.

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

Proof.

$V_K\in C(I)$ is quasinilpotent so applying (1) with $\lambda =1$,

$${(e-V_K)}^{-1}=\underset{n\to \mathrm{\infty }}{lim}\sum _{j=0}^n{V}_K^j\in A.$$

Then

$${(e-V_K)}^{-1}{u}_0=\left(\underset{n\to \mathrm{\infty }}{lim}\sum _{j=0}^n{V}_K^j\right)u_0=\underset{n\to \mathrm{\infty }}{lim}(V_K^j{u}_0)=\sum _{n=0}^{\mathrm{\infty }}{V}_K^n{u}_0.$$

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 $x\in I$,

$u(x)$ $=u_0(x)+{\displaystyle {\int }_0^x}K(x,y)u(y)𝑑y.$

∎

Theorem 8.

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

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

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

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

∎

4 Gronwall’s inequality

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

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

which implies

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

The Lebesgue differentiation theorem states that for almost all $x\in I$, $x$ is a Lebesgue point of $f$. Let

$$F(x)={\int }_0^{x}f(y)𝑑y,x\in I,$$

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.

Theorem 9 (Gronwall’s inequality).

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

$$y(t)\le f(t)+{\int }_0^{t}g(s)y(s)𝑑s,t\in I,$$

then

$$y(t)\le f(t)+{\int }_0^{t}f(s)g(s)\exp \left({\int }_s^{t}g(u)𝑑u\right)𝑑s,t\in I.$$

If $f$ is increasing then

$$y(t)\le f(t)\exp \left({\int }_0^{t}g(s)𝑑s\right),t\in I.$$

Proof.

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

$$Z(t)={\int }_0^{t}z(s)𝑑s,t\in I.$$

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

$$Z^{\prime }(t)=z(t)=g(t)y(t)\le g(t)\left(f(t)+{\int }_0^{t}g(s)y(s)𝑑s\right)=g(t)f(t)+g(t)Z(t).$$

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

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

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

${\left[\exp \left(-{\displaystyle {\int }_0^s}g(u)𝑑u\right)Z(s)\right]}^{\prime }$ $=\exp \left(-{\displaystyle {\int }_0^s}g(u)𝑑u\right)\left[Z^{\prime }(s)-g(t)Z(s)\right].$

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

But

${\displaystyle {\int }_0^t}{\left[\exp \left(-{\displaystyle {\int }_0^s}g(u)𝑑u\right)Z(s)\right]}^{\prime }𝑑s$ $={\left[\exp \left(-{\displaystyle {\int }_0^s}g(u)𝑑u\right)Z(s)\right]|}_0^t$

$=\exp \left(-{\displaystyle {\int }_0^t}g(u)𝑑u\right)Z(t).$

So

$$\exp \left(-{\int }_0^{t}g(u)𝑑u\right)Z(t)\le {\int }_0^{t}g(s)f(s)\exp \left(-{\int }_0^{s}g(u)𝑑u\right)𝑑s.$$

Therefore,

$y(t)$ $\le f(t)+{\displaystyle {\int }_0^t}g(s)y(s)𝑑s$

$=f(t)+Z(t)$

$\le f(t)+\exp \left({\displaystyle {\int }_0^t}g(u)𝑑u\right){\displaystyle {\int }_0^t}g(s)f(s)\exp \left(-{\displaystyle {\int }_0^s}g(u)𝑑u\right)𝑑s$

$=f(t)+{\displaystyle {\int }_0^t}g(s)f(s)\exp \left({\displaystyle {\int }_0^t}g(u)𝑑u-{\displaystyle {\int }_0^s}g(u)𝑑u\right)𝑑s$

$=f(t)+{\displaystyle {\int }_0^t}g(s)f(s)\exp \left({\displaystyle {\int }_s^t}g(u)𝑑u\right)𝑑s.$

Suppose that $f$ is increasing. Let

$$G(s)={\int }_0^{s}g(u)𝑑u,s\in I.$$

For $t\in I$,

$y(t)$ $\le f(t)+{\displaystyle {\int }_0^t}g(s)f(s)\exp \left({\displaystyle {\int }_s^t}g(u)𝑑u\right)𝑑s$

$\le f(t)+{\displaystyle {\int }_0^t}g(s)f(t)\exp \left({\displaystyle {\int }_s^t}g(u)𝑑u\right)𝑑s$

$=f(t)\left[1+{\displaystyle {\int }_0^t}g(s)\exp \left({\displaystyle {\int }_s^t}g(u)𝑑u\right)𝑑s\right]$

$=f(t)\left[1+{\displaystyle {\int }_0^t}g(s)e^{G(t)-G(s)}𝑑s\right]$

$=f(t)\left[1+e^{G(t)}{\displaystyle {\int }_0^t}g(s)e^{-G(s)}𝑑s\right].$

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

$$y(t)\le f(t)\left[1+\frac{1}{H(t)}{\int }_0^{t}g(s)H(s)𝑑s\right].$$

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

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

Hence

$y(t)$ $\le f(t)\left[1-{\displaystyle \frac{1}{H(t)}}{\displaystyle {\int }_0^t}{H}^{\prime }(s)𝑑s\right]$

$=f(t)\left[1-{\displaystyle \frac{1}{H(t)}}\left[H(t)-H(0)\right]\right]$

$=f(t)\left[1-1+{\displaystyle \frac{H(0)}{H(t)}}\right]$

$=f(t)e^{G(t)}$

$=f(t)\exp \left({\displaystyle {\int }_0^t}g(u)𝑑u\right).$

∎

Let $K(x,y)=(x-y)Q(y)$. Let $u={\sum }_{n=0}^{\mathrm{\infty }}{V}_K^n{u}_0\in C(I)$. 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

$$|u(x)|\le x+{\int }_0^x|x-y||Q(y)||u(y)|𝑑y\le x+{\int }_0^x|Q(y)||u(y)|𝑑y.$$

Applying Gronwall’s inequality we get

$$|u(x)|\le x\exp \left({\int }_0^x|Q(y)|𝑑y\right),x\in I.$$ (2)

5 The spectral theorem for positive compact operators

The following is the spectral theorem for positive compact operators.

Theorem 10 (Spectral theorem for positive compact operators).

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

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 $\mathrm{\Phi }\subset H$ and ${\lambda }_{\varphi }>0$ for $\varphi \in \mathrm{\Phi }$ such that $\mathrm{\Phi }$ is an orthonormal basis for $H$, $T\varphi ={\lambda }_{\varphi }\varphi$ for $\varphi \in \mathrm{\Phi }$, and the unique limit point of $\{{\lambda }_{\varphi }:\varphi \in \mathrm{\Phi }\}$ is $0$. Let $\mathrm{\Phi }=\{{\varphi }_n:n\ge 1\}$, ${\varphi }_n\ne {\varphi }_m$ for $n\ge m$, such that $n\ge m$ implies ${\lambda }_{{\varphi }_n}\le {\lambda }_{{\varphi }_m}$. Let ${\lambda }_n={\lambda }_{{\varphi }_n}$. Then ${\lambda }_n\downarrow 0$. Summarizing, there is an orthonormal basis $\{{\varphi }_n:n\ge 1\}$ for $H$ and ${\lambda }_n>0$ such that $T{\varphi }_n={\lambda }_n{\varphi }_n$ for $n\ge 1$ and ${\lambda }_n\downarrow 0$.

6 $Q>0$, Green’s function for $L_Q$

Suppose $Q\in C(I)$ with $Q(x)>0$ for . 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

$$u=\sum _{n=0}^{\mathrm{\infty }}{V}_K^n{u}_0\in C(I).$$

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

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

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

$$V_{K}f(x)={\int }_0^x(x-y)Q(y)f(y)𝑑y>0.$$

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

$$u(x)=\sum _{n=0}^{\mathrm{\infty }}(V_K^n{u}_0)(x)>0.$$

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 and $u(y)>0$ for ,

Using $u(x)=x+{\int }_0^x(x-y)Q(y)u(y)𝑑y$ and $Q>0$ we get

Let $u_1(x)=u(x)$ and $u_2(x)=u(1-x)$. Then

$$L_Q{u}_1=0,u_1(0)=0,u_1^{\prime }(0)=1$$

and

$$L_Q{u}_2=0,u_2(1)=0,u_2^{\prime }(1)=-1.$$

A fortiori,

and as $u_2^{\prime }(x)=-u^{\prime }(1-x)$,

For 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 $u_1^{\prime }{u}_2>0$. so $-u_1{u}_2^{\prime }>0$, hence $W>0$.

$W^{\prime }$ $={(u_1^{\prime }{u}_2-u_1{u}_2^{\prime })}^{\prime }$

$=u_1^{\prime \prime }{u}_2+u_1^{\prime }{u}_2^{\prime }-u_1^{\prime }{u}_2^{\prime }-u_1{u}_2^{\prime \prime }$

$=u_1^{\prime \prime }{u}_2-u_1{u}_2^{\prime \prime }$

$=(Q{u}_1)u_2-u_1(Q{u}_2)$

$=0.$

Therefore there is some $W_0>0$ such that $W(x)=W_0$ for all .

Define

$$G(x,y)=\frac{u_1(x\wedge y)u_2(x\vee y)}{W_0},(x,y)\in I\times I.$$

$x\wedge y=\min (x,y)$, $x\vee y=\max (x,y)$. 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 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

$$⟨f,g⟩={\int }_{I}f\overline{g}𝑑\mu ,f,g\in L^2(I).$$

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.

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

$⟨T_{Q}g,f⟩$ $={\displaystyle {\int }_I}(T_{Q}g)(x)\overline{f(x)}𝑑x$

$={\displaystyle {\int }_I}\left({\displaystyle {\int }_I}G(x,y)g(y)𝑑y\right)\overline{f(x)}𝑑x$

$={\displaystyle {\int }_I}g(y)\overline{\left({\displaystyle {\int }_I}G(y,x)f(x)𝑑x\right)}𝑑y$

$={\displaystyle {\int }_I}g(y)\overline{(T_{Q}f)(y)}𝑑y$

$=⟨g,T_{Q}f⟩.$

Therefore $T_Q:L^2(I)\to L^2(I)$ is self-adjoint.

We now establish properties of $T_Q$. Let

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

Lemma 11.

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

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

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

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

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

Proof.

For $x\in I$,

$f(x)$ $={\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(x\wedge y)u_2(x\vee y)}{W_0}}g(y)𝑑y+{\displaystyle {\int }_x^1}{\displaystyle \frac{u_1(x\wedge y)u_2(x\vee y)}{W_0}}g(y)𝑑y$

$={\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)u_2(x)}{W_0}}g(y)𝑑y+{\displaystyle {\int }_x^1}{\displaystyle \frac{u_1(x)u_2(y)}{W_0}}g(y)𝑑y$

$=u_2(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_1(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y.$

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

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

$f^{\prime }(x)$ $=u_2^{\prime }(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_2(x){\displaystyle \frac{u_1(x)g(x)}{W_0}}$

$+u_1^{\prime }(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y-u_1(x){\displaystyle \frac{u_2(x)g(x)}{W_0}}$

$=u_2^{\prime }(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_1^{\prime }(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y.$

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

$f^{\prime \prime }(x)$ $=u_2^{\prime \prime }(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_2^{\prime }(x){\displaystyle \frac{u_1(x)g(x)}{W_0}}$

$+u_1^{\prime \prime }(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y-u_1^{\prime }(x){\displaystyle \frac{u_2(x)g(x)}{W_0}}$

$=u_2^{\prime \prime }(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_1^{\prime \prime }(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y-{\displaystyle \frac{W(x)g(x)}{W_0}}$

$=u_2^{\prime \prime }(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+u_1^{\prime \prime }(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y-g(x).$

Because $g\in C(I)$ it follows that $f^{\prime \prime }\in C(I)$, i.e. $f\in C^2(I)$. Furthermore, because $u_1^{\prime \prime }=Q{u}_1$ and $u_2^{\prime \prime }=Q{u}_2$,

$f^{\prime \prime }(x)$ $=Q(x)u_2(x){\displaystyle {\int }_0^x}{\displaystyle \frac{u_1(y)g(y)}{W_0}}𝑑y+Q(x)u_1(x){\displaystyle {\int }_x^1}{\displaystyle \frac{u_2(y)g(y)}{W_0}}𝑑y-g(x)$

$=Q(x)f(x)-g(x).$

∎

We now establish more facts about $T_Q$.

Lemma 12.

Let $Q\mathrm{\in }C\mathit{}\mathrm{(}I\mathrm{)}$, $Q\mathit{}\mathrm{(}x\mathrm{)}\mathrm{>}\mathrm{0}$ for .

1. If $f_1,f_2\in N^2(I)$ then

   $${\int }_I{f}_1{L}_Q{f}_2𝑑x={\int }_I(f_1^{\prime }{f}_2^{\prime }+Q{f}_1{f}_2)𝑑x.$$

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

3. If $f\in N^2(I)$ then $f=T_Q{L}_{Q}f$.

4. $T_Q\ge 0$.

5. $\ker T_Q=0$.

Proof.

First, doing integration by parts,

${\displaystyle {\int }_I}{f}_1(-f_2^{\prime \prime }+Q{f}_2)𝑑x$ $=-{\displaystyle {\int }_{\partial I}}{f}_1{f}_2^{\prime }+{\displaystyle {\int }_I}{f}_1^{\prime }{f}_2^{\prime }𝑑x+{\displaystyle {\int }_I}Q{f}_1{f}_2𝑑x$

$={\displaystyle {\int }_I}{f}_1^{\prime }{f}_2^{\prime }𝑑x+{\displaystyle {\int }_I}Q{f}_1{f}_2𝑑x$

$={\displaystyle {\int }_I}(f_1^{\prime }{f}_2^{\prime }+Q{f}_1{f}_2)𝑑x.$

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 , it follows that $|f|=0$ almost everywhere. But $f$ is continuous so $f=0$. For $f=f_1+i{f}_2$, 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

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

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,

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

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 $f=f_1+i{f}_2$, if $f(0)=0$, $f(1)=0$ and $g=-f^{\prime \prime }+Qf$, let $g=g_1+i{g}_2$. As $Q$ is real-valued we get $g_1=-f_1^{\prime \prime }+Q{f}_1$ and $g_2=-f_2^{\prime \prime }+Q{f}_2$. Then $f_1=T_Q{g}_1$ and $f_2=T_Q{g}_2$. Thus

$$f=f_1+i{f}_2=T_Q{g}_1+i{T}_Q{g}_2=T_Q(g_1+i{g}_2)=T_{Q}g.$$

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

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

Then using the above,

$⟨g,T_{Q}g⟩$ $=⟨-f^{\prime \prime }+Qf,f⟩$

$={\displaystyle {\int }_I}(-f^{\prime \prime }+Qf)\overline{f}𝑑x$

$={\displaystyle {\int }_I}({\overline{f}}^{\prime }{f}^{\prime }+Q\overline{f}f)𝑑x$

$={\displaystyle {\int }_I}({|f^{\prime }|}^2+Q{|f|}^2)𝑑x.$

Because $Q\ge 0$ we have $⟨g,T_{Q}g⟩\ge 0$. For $g\in L^2(I)$ let $g_n\in C(I)$ with ${\parallel g_n-g\parallel }_{L^2}\to 0$. Then $⟨g_n,T_Q{g}_n⟩\to ⟨g,T_{Q}g⟩$ as $n\to \mathrm{\infty }$, and because $⟨g_n,T_Q{g}_n⟩\ge 0$ it follows that $⟨g,T_{Q}g⟩\ge 0$. Therefore $T_Q\ge 0$, 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 $⟨f,T_Q^{*}g⟩=⟨T_{Q}f,g⟩=0$. Hence $\ker (T_Q)⟂\mathrm{Ran}(T_Q^{*})$. But $T_Q$ is self-adjoint which implies that $\ker (T_Q)⟂\mathrm{Ran}(T_Q)$. 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.

Theorem 13 (Sturm-Liouville theorem).

Let $Q\mathrm{\in }C\mathit{}\mathrm{(}I\mathrm{)}$, $Q\mathit{}\mathrm{(}x\mathrm{)}\mathrm{>}\mathrm{0}$ for . There is an orthonormal basis $\mathrm{\{}{u}_n\mathrm{:}n\mathrm{\ge }\mathrm{1}\mathrm{\}}\mathrm{\subset }{N}^{\mathrm{2}}\mathit{}\mathrm{(}I\mathrm{)}$ for $L^{\mathrm{2}}\mathit{}\mathrm{(}I\mathrm{)}$ and ${\lambda }_n\mathrm{>}\mathrm{0}$, for and ${\lambda }_n\mathrm{\to }\mathrm{\infty }$, such that

$$L_Q{u}_n={\lambda }_n{u}_n,n\ge 1.$$

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 $\{{\varphi }_n:n\ge 1\}$ for $L^2(I)$ and ${\gamma }_n>0$ such that $T_Q{\varphi }_n={\gamma }_n{\varphi }_n$ for $n\ge 1$ and ${\gamma }_n\downarrow 0$. By Lemma 11, $T_Q{\varphi }_n\in N^0(I)$. Let

$$u_n=\frac{1}{{\gamma }_n}{T}_Q{\varphi }_n\in N^0(I).$$

Because $T_Q{\varphi }_n={\gamma }_n{\varphi }_n$ we have $u_n={\varphi }_n$ 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 ${\lambda }_n>0$, ${\lambda }_m\le {\lambda }_n$ for $m\le n$, ${\lambda }_n\to \mathrm{\infty }$, and

$$L_Q{u}_n={\lambda }_n{u}_n,n\ge 1.$$

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

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

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 $W(x)\ne 0$ 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 $n\ne m$. 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 $\{u_n:n\ge 1\}$ is an orthonormal set. Therefore $m\ne n$ implies that ${\lambda }_m\ne {\lambda }_n$. ∎

8 Other results in Sturm-Liouville theory