Along with the problem \({\mathcal {A}}={\mathcal {A}}(1,M,g,v)\), we consider the problem \({\mathcal {A}^{*}}={\mathcal {A}}(1,M,u,w)\), where \(w(x)=g(\pi -x)\) and \(u(x)=v(\pi -x)\). Let \(y=u_{1}(x,\lambda)\) be the solution of the Cauchy problem

$$ \ell_{1}^{*} y:=iy'+ \int_{0}^{x} M(x-t)y(t)\,dt+u(x)=\lambda y,\quad 0< x< \pi, y(0)=0. $$

(21)

According to Lemma 2, the following representation holds:

$$ u_{1}(x,\lambda)=- \int_{0}^{x} S_{1}(x-t,\lambda)u(t)\,dt. $$

(22)

### Lemma 7

*Let*
\(n=1\). *Then the following relation holds*:

$$ \hat{\Delta }(\lambda)= - \int_{0}^{\pi }N(x){\mathcal {M}}(x,\lambda)\,dx, $$

(23)

*where*

$$\begin{aligned}& \begin{aligned} &N(x)= \int_{x}^{\pi }\hat{M}(\pi -t)\,dt \int_{0}^{t-x}u(\tau)g(t-x- \tau)\,d\tau, \\ &{\mathcal {M}}(x,\lambda)= \int_{0}^{x}S_{1}(x-t, \lambda) \tilde{S}_{1}(t,\lambda)\,dt. \end{aligned} \end{aligned}$$

(24)

### Proof

According to (5) and (21), we have

$$\begin{aligned} 0 =&u_{1}(\pi -x,\lambda)\tilde{\ell }_{1} \tilde{g}_{1}(x,\lambda) - \tilde{g}_{1}(x,\lambda) \ell_{1}^{*}u_{1}(\pi -x,\lambda) \\ =& i \bigl(u_{1}( \pi -x,\lambda)\tilde{g}_{1}(x,\lambda) \bigr)' \\ &{}+u_{1}(\pi -x,\lambda) \biggl( \int_{0}^{x}\tilde{M}(x-t)\tilde{g}_{1}(t, \lambda)\,dt + g(x) \biggr) \\ &{} - \tilde{g}_{1}(x,\lambda) \biggl( \int_{0} ^{\pi -x}\tilde{M}(\pi -x-t)u_{1}(t, \lambda)\,dt + v(x) \biggr). \end{aligned}$$

Integrating this from 0 to *π* and changing the order of integration, we arrive at

$$ \Delta^{*}(\lambda)-\tilde{\Delta }(\lambda)= - \int_{0}^{\pi } \hat{M}(\pi -x)\,dx \int_{0}^{x} u_{1}(t,\lambda) \tilde{g}_{1}(x-t, \lambda)\,dt, $$

(25)

where

$$\Delta^{*}(\lambda)=1- \int_{0}^{\pi }w(x)u_{1}(x,\lambda)\,dx. $$

Formula (25), in particular, gives \(\Delta^{*}(\lambda)=\Delta (\lambda)\) and, hence,

$$ \hat{\Delta }(\lambda)= - \int_{0}^{\pi }\hat{M}(\pi -x)\,dx \int_{0} ^{x} u_{1}(t,\lambda) \tilde{g}_{1}(x-t,\lambda)\,dt. $$

(26)

Substituting (8) for \(n=1\) and (22) into (26) and changing the order of integration, we arrive at (23) and (24). □

### Lemma 8

*The following representation holds*:

$$ {\mathcal {M}}(x,\lambda)= - \biggl(x\exp (-i\lambda x)+ \int_{0}^{x} K(x,t) \exp (-i\lambda t)\,dt \biggr), \quad 0 \le x\le \pi, $$

(27)

*where*
\(\vert K(x,t)\vert \le f(x-t)\)
*with some*
\(f(x)\in L_{2}(0,\pi)\).

### Proof

Substituting (12) for \(n=1\) into the second formula in (24), we get

$${\mathcal {M}}_{1}(x,\lambda)=- \Biggl(x\exp (-i\lambda x)+ \sum _{j=1}^{3} {\mathcal {M}}_{j}(x,\lambda) \Biggr), $$

where, putting \(Q(x,t)=P(x,x-t)\), we have

$$\begin{aligned}& {\mathcal {M}}_{1}(x,\lambda)= \int_{0}^{x}\exp (-i\lambda t)\,dt \int_{0} ^{x-t} Q(x-t,\tau)\exp (-i\lambda \tau)\,d\tau, \\& {\mathcal {M}}_{2}(x,\lambda)= \int_{0}^{x}\exp (-i\lambda t)\,dt \int_{0} ^{x-t} \tilde{Q}(x-t,\tau)\exp (-i\lambda \tau)\,d\tau, \\& {\mathcal {M}}_{3}(x,\lambda)= \int_{0}^{x}\,dt \int_{0}^{t} Q(t,\tau) \exp (-i\lambda \tau)\,d\tau \int_{0}^{x-t} \tilde{Q}(x-t,\xi) \exp (-i\lambda \xi)\,d\xi. \end{aligned}$$

Combining the exponentials and changing the order of integration in the last three formulae, we arrive at (27) with

$$\begin{aligned} K(x,t) =& \int_{0}^{t} Q(x-\tau,t-\tau)\,d\tau + \int_{0}^{t}\tilde{Q}(x- \tau,t-\tau)\,d\tau \\ &{} + \int_{0}^{t}\,d\tau \int_{\tau }^{x+\tau -t} Q( \xi,\tau)\tilde{Q}(x-\xi,t-\tau)\,d\xi, \end{aligned}$$

which along with (13) finishes the proof. □

Now we are in a position to give the proof of Theorem 1. First, let \(n=1\). Changing the order of integration in (24), we get

$$ N(x)= \int_{x}^{\pi }v(t)\,dt \int_{0}^{t-x} \hat{M}(t-x-\tau)g(\tau)\,d\tau. $$

(28)

Without loss of generality, we assume that \(g(x)=0\) a.e. on \((0,a)\) and \(v(x)=0\) a.e. on \((b,\pi)\). Otherwise, the points \(a>0\) and \(b<\pi \) could be shifted closer to the points 0 and *π*, respectively, which would make Theorem 1 even stronger. Denote

$$g_{1}(x):=g(x+a),\quad x\in (0,\pi -a),\qquad v_{1}(x):=v(b-x),\quad x \in (0,b). $$

Then, by virtue of (3), we get

$$ \int_{0}^{\varepsilon }\bigl\vert g_{1}(x)\bigr\vert \,dx>0,\qquad \int_{0}^{\varepsilon }\bigl\vert v_{1}(x)\bigr\vert \,dx>0, \quad \varepsilon >0. $$

(29)

According to (28), we have \(N(x)=0\) on \((b,\pi)\) and

$$\begin{aligned} N(b-x) =& \int_{0}^{x} v_{1}(t)\,dt \int_{0}^{x-t} \hat{M}(x-t-\tau)g( \tau)\,d\tau \\ =& \int_{0}^{x} g(t)\,dt \int_{0}^{x-t} \hat{M}(x-t- \tau)v_{1}( \tau)\,d\tau \end{aligned}$$

for \(x\in (0,b)\). Hence, \(N(b-x)=0\) for \(x\in (0,a)\), i.e., \(N(x)=0\) on \((b-a,\pi)\), and

$$\begin{aligned}& \begin{aligned} &N(b-a-x)= \int_{0}^{x} g_{1}(t)F(x-t)\,dt, \\ & F(x)= \int_{0}^{x} \hat{M}(x-t)v_{1}(t)\,dt,\quad x\in (0,b-a). \end{aligned} \end{aligned}$$

(30)

Further, substituting (27) into (23) and changing the order of integration, we obtain

$$\hat{\Delta }(\lambda)= \int_{0}^{\pi } \biggl(xN(x)+ \int_{x}^{\pi }K(t,x) N(t)\,dt \biggr)\exp (-i\lambda x)\,dx. $$

According to Lemma 5, the coincidence of the spectra gives

$$xN(x)+ \int_{x}^{\pi }K(t,x) N(t)\,dt=0\quad \text{a.e. on } (0, \pi), $$

which implies that \(N(x)=0\) a.e. on \((0,\pi)\). By virtue of the first inequality in (29) and (30) along with the Titchmarsh theorem (see [35]), we obtain \(F(x)=0\) a.e. on \((0,b-a)\). Applying the Titchmarsh theorem to the second equality in (30) and taking into account the second inequality in (29), we arrive at \(\hat{M}(x)=0\) a.e. on \((0,b-a)\). Thus, part (i) of Theorem 1 is proven for \(n=1\).

Let \(n>1\). Along with \({\mathcal {A}}\) we consider the problem \({\mathcal {A}} _{1}={\mathcal {A}}(1,M_{1},g,v)\), where the function \(M_{1}(x)\) is connected with \(M(x)\) by relation (11). According to Lemma 4, the characteristic function \(\Delta_{1}(\lambda)\) of \({\mathcal {A}}_{1}\) has the form

$$ \Delta_{1}(\lambda)=1 -i \int_{0}^{\pi }\mu (x)\exp (-i\lambda x)\,dx, $$

(31)

with the same function \(\mu (x)\) that appears in representation (14) for the characteristic function \(\Delta (\lambda)\) of \({\mathcal {A}}\).

Further, if \(\gamma =\tilde{\gamma }\) and \(\{\lambda_{k}\}_{k\ge 1}= \{\tilde{\lambda }_{k}\}_{k\ge 1}\), then formula (16) implies \(\Delta (\lambda)\equiv \tilde{\Delta }(\lambda)\). By virtue of (14), this infers \(\mu (x)=\tilde{\mu }(x)\) and, hence, (31) gives \(\Delta_{1}(\lambda)\equiv \tilde{\Delta }_{1}(\lambda)\). According to part (i) of Theorem 1 for \(n=1\), we get \(M_{1}(x)= \tilde{M}_{1}(x)\) a.e. on \((0,b-a)\) and, by virtue of (11), \(M(x)=\tilde{M}(x)\) a.e. on \((0,b-a)\). Thus, part (i) for \(n=2\) and part (ii) of Theorem 1 are proven.

We actually have established that \(\Delta (\lambda)\equiv \tilde{\Delta }(\lambda)\) if and only if \(\hat{M}(x)=0\) a.e. on \((0,b-a)\), which proves part (iii) of Theorem 1.

For the proof of Theorem 2, it remains to note that under its hypothesis, according to Lemma 6, the specification of the spectrum uniquely determines *γ*.