In this section we collect some preliminary results that will be used below.
We denote the usual norm in \(L^{1}(0,T)\) by \(\|\cdot \|_{L^{1}}\). Let \(X:=C[0,T]\) be the Banach space endowed with the uniform norm \(\|\cdot \|_{\infty }\), \(Y:=C^{1}[0,T]\) be the Banach space equipped with the norm \(\|u\|_{C^{1}}=\|u\|_{\infty }+\|u'\|_{\infty }\), the corresponding open ball of center at 0 and radius r is denoted by \(B_{r}\).
Definition 2.1
Let \(\omega : X\rightarrow \mathbb{R}\) be a functional. ω is increasing if
$$ x,y\in X,\quad x(t)< y(t) \quad \text{for } t\in [0,T], \quad \text{then } \omega (x) \leq \omega (y). $$
For each \(\omega : X\rightarrow \mathbb{R}\), \(\operatorname{Im}(\omega )\) denotes the range of ω.
Set \(\mathcal{A}=\{\omega \mid \omega :X \rightarrow \mathbb{R} \text{ is continuous and increasing}\}\), \(\mathcal{A}_{0}=\{\omega \mid \omega \in \mathcal{A}, \omega (0)=0\}\).
Remark 2.2
Conspicuously, \(\min \{u(t) \mid t\in [0,T]\}\) and \(\max \{u(t)\mid t\in [0,T]\}\) belong to \(\mathcal{A}\). If we take
$$ \omega (u)=\min \bigl\{ u(t) \mid t\in [0,T] \bigr\} , $$
then the boundary condition (1.2) is equal to
$$ \omega (u)=A, \max \bigl\{ u(t) \mid t\in [0,T] \bigr\} -\min \bigl\{ u(t) \mid t \in [0,T] \bigr\} =B-A. $$
(2.1)
So, in the rest part of the paper we only deal with (1.1) and (2.1).
Lemma 2.3
([20, Lemma 4])
Let
\(\omega \in \mathcal{A}\), \(k \in [0,1]\)and
\(u\in X\), the equality
\(\omega (u)-k\omega (-u)=0\)is satisfied. Then there exists a
\(\delta \in [0,T]\)such that
\(u(\delta )=0\).
Lemma 2.4
([20, Lemma 5])
Let
\(\omega \in \mathcal{A}\), \(h \in \operatorname{Im}(\omega )\). Then there exists a unique
\(k\in X\)such that
\(\omega (k)=h\).
Lemma 2.5
(Bihari lemma, [19, Lemma 2.1]; [20, Lemma 1])
Let
\(p:[0,+\infty )\rightarrow (0,+\infty )\)be a nondecreasing continuous function, \(P:[0,+\infty )\rightarrow [0,+\infty )\)be defined by
\(P(u)=\int _{0}^{u}\frac{dt}{p(t)}\)and let
\(b\in [c,d]\subset \mathbb{R}\). If
\(v\in X\)satisfies the inequality
$$ \bigl\vert v(t) \bigr\vert \leq \biggl\vert \int _{b}^{t}p \bigl( \bigl\vert v(s) \bigr\vert \bigr) \biggr\vert \,ds,\quad \textit{for } t\in [c,d], $$
then
$$ \bigl\vert v(t) \bigr\vert \leq P^{-1}(b-t), \quad \textit{for } t \in [c,b], $$
provided
\(\lim_{u\rightarrow \infty } P(u)>b-c\), and
$$ \bigl\vert v(t) \bigr\vert \leq P^{-1}(t-b),\quad \textit{for } t \in [b,d], $$
provided
\(\lim_{u\rightarrow \infty }P(u)>d-b\). Here
\(P^{-1}\)denotes the inverse function toP.
As in [5], we define the function \(\psi : X\rightarrow \mathbb{R}\) by the formula
$$ \psi (u)=\max \biggl\{ \int _{m}^{n}u(s)\,ds \Bigm| m,n\in [0,T], m\leq n \biggr\} . $$
(2.2)
Lemma 2.6
([5])
For all
\(u\in Y\), the functionalψis continuous and
$$ \max \bigl\{ u(t) \mid t\in [0,T] \bigr\} -\min \bigl\{ u(t) \mid t\in [0,T] \bigr\} =\max \bigl\{ \psi \bigl(u' \bigr), \psi \bigl(-u' \bigr) \bigr\} . $$
Lemma 2.7
Suppose thatuis a solution of (1.1) on
\([0,T]\). Then
$$ \min \bigl\{ \psi \bigl(u' \bigr), \psi \bigl(-u' \bigr) \bigr\} \leq \frac{T}{2}\phi ^{-1} \biggl(P^{-1} \biggl( \frac{T}{2} \biggr) \biggr), $$
(2.3)
where
\(P^{-1}\)denotes the inverse function to
$$ P(u)= \int _{0}^{u}\frac{ds}{f(\phi ^{-1}(s))}. $$
Proof
Set
$$ C_{+}= \bigl\{ t\mid u'(t)>0, t\in (0,T) \bigr\} ,\qquad C_{-}= \bigl\{ t\mid u'(t)< 0, t\in (0,T) \bigr\} . $$
Let \(\mu (C_{+})\) and \(\mu (C_{-})\) be the Lebesgue measure of \(C_{+}\), \(C_{-}\), respectively.
If \(C_{+}=\emptyset\) (resp. \(C_{-}=\emptyset \)), then \(\psi (u')=0\) (resp. \(\psi (-u')=0\)) and (2.3) is clearly established. Assume \(C_{+}\neq \emptyset \) and \(C_{-}\neq \emptyset \). \(u'\in X\), \(C_{+}\), \(C_{-}\) are open subsets of \([0,T]\) and therefore \(C_{+}\) (resp. \(C_{-}\)) is a union of at most countable set of disjoint open intervals \((a_{i},b_{i})\), \(i\in I_{+}\subset \mathbb{N}\) (resp. \((c _{j},d_{j})\), \(j\in I_{-}\subset \mathbb{N}\)) without common elements, i.e.
$$ C_{+}=\bigcup_{i\in I_{+}}(a_{i},b_{i}),\qquad C_{-}=\bigcup_{j\in I_{-}}(c_{j},d_{j}). $$
Of course, for any \(i\in I_{+}\), \(u'(a_{i})\neq 0\) or \(u'(b_{i}) \neq 0\) (resp. \(u'(c_{j})\neq 0\) or \(u'(d_{j})\neq 0\) for any \(j \in I_{-}\)) imply \(a_{i}=0\) or \(b_{i}=T\) (resp. \(c_{j}=0\) or \(d _{j}=T\)). Furthermore, \(C_{+}\neq (0,T)\), since in the opposite case \(C_{-}=\emptyset \), which makes a contradiction. Similarly, \(C_{-}\neq (0,T)\).
By the inequality \(\mu (C_{+})+\mu (C_{-})\leq T\), it is easy to see that
$$ \min \bigl\{ \mu (C_{+}),\mu (C_{-}) \bigr\} \leq \frac{T}{2}. $$
(2.4)
Next we prove the inequality
$$ \psi \bigl(u' \bigr)\leq \mu (C_{+})\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(b_{i}-a_{i}) \bigr)\mid i\in I_{+} \bigr\} . $$
(2.5)
Fix \(i\in I_{+}\), let \(u'(\eta )=0\), \(\eta \in \{a_{i},b_{i}\}\). Combining (1.1) with \(\phi (0)=0\), we have
$$ \phi \bigl(u'(t) \bigr)= \int _{\eta }^{t}(Fu) (s)\,ds,\quad t\in [a_{i},b_{i}]. $$
For \(t\in [a_{i},b_{i}]\), \(u'(t)\geq 0\). Since ϕ is an increasing homeomorphism and because of (H1), we get
$$ 0\leq \phi \bigl(u'(t) \bigr)\leq \biggl\vert \int _{\eta }^{t} \bigl\vert (Fu) (s) \bigr\vert \,ds \biggr\vert \leq \biggl\vert \int _{\eta }^{t}f \bigl(u'(s) \bigr) \,ds \biggr\vert = \biggl\vert \int _{\eta }^{t}f\bigl(\phi ^{-1} \bigl(\phi \bigl(u'(s) \bigr) \bigr)\,ds \biggr\vert . $$
(2.6)
From Lemma 2.5 with \(b=\eta \), \(c=a_{i}\), \(d=b_{i}\), \(v(s)=\phi (u'(s))\) and \(p(v)=f(\phi ^{-1}(v))\), it is not difficult to see that
$$ \phi \bigl(u'(t) \bigr)\leq P^{-1} \bigl( \vert \eta -t \vert \bigr), \quad t\in [a_{i},b_{i}]. $$
Subsequently, \(0\leq u'(t)\leq \phi ^{-1} (P^{-1}(b_{i}-a_{i}) )\) for \(t\in [a_{i},b_{i}]\), \(i\in I_{+}\). Thus
$$ \int _{a_{i}}^{b_{i}}u'(s)\,ds\leq (b_{i}-a_{i})\phi ^{-1} \bigl(P^{-1}(b _{i}-a_{i}) \bigr). $$
(2.7)
Moreover,
$$\begin{aligned} \psi \bigl(u' \bigr) \leq & \int _{C_{+}}u'(t)\,dt=\sum _{i\in I_{+}} \int _{a_{i}}^{b_{i}}u'(t)\,dt \\ \leq &\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(b_{i}-a_{i}) \bigr)\mid i\in I_{+} \bigr\} \sum_{i\in I_{+}}(b_{i}-a_{i}) \\ \leq &\mu (C_{+})\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(b_{i}-a_{i}) \bigr)\mid i \in I_{+} \bigr\} . \end{aligned}$$
As a consequence, (2.5) is satisfied.
Next, we will show that
$$ \psi \bigl(-u' \bigr)\leq \mu (C_{-})\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(d_{j}-c_{j}) \bigr)\mid j\in I_{-} \bigr\} . $$
(2.8)
Fix \(j\in I_{-}\), let \(u'(\zeta )=0\), \(\zeta \in \{c_{j},d_{j}\}\). Together (1.1) with \(\phi (0)=0\), which implies
$$ \phi \bigl(u'(t) \bigr)= \int _{\zeta }^{t}(Fu) (s)\,ds, \quad t\in [c_{j},d_{j}]. $$
We have \(u'(t)\leq 0\) on \([c_{j},d_{j}]\). Combining the fact that ϕ is an odd increasing homeomorphism and (H1), we obtain
$$ 0\leq -\phi \bigl(u'(t) \bigr)\leq \biggl\vert \int _{\zeta }^{t} \bigl\vert (Fu) (s) \bigr\vert \,ds \biggr\vert \leq \biggl\vert \int _{\zeta }^{t}f \bigl(-u'(s) \bigr) \,ds \biggr\vert . $$
(2.9)
Thus
$$ \phi \bigl( \bigl\vert u'(t) \bigr\vert \bigr)=-\phi \bigl(u'(t) \bigr)\leq \biggl\vert \int _{\zeta }^{t}f\bigl(\phi ^{-1} \bigl( \phi \bigl( \bigl\vert u'(s) \bigr\vert \bigr) \bigr)\,ds \biggr\vert . $$
(2.10)
From Lemma 2.5 with \(b=\zeta \), \(c=c_{j}\), \(d=d_{j}\), \(v(s)=\phi (|u'(s)|)\) and \(p(v)=f(\phi ^{-1}(v))\), it is easy to verify that
$$ \phi \bigl( \bigl\vert u'(t) \bigr\vert \bigr)\leq P^{-1} \bigl( \vert t-\zeta \vert \bigr),\quad t\in [c_{j},d_{j}]. $$
Hence, \(0\leq -u'(t)\leq \phi ^{-1} (P^{-1}(d_{j}-c_{j}) )\) for \(t\in [c_{j},d_{j}]\), \(j\in I_{-}\). So
$$ -\int _{c_{j}}^{d_{j}}u'(t)\,dt\leq (d_{j}-c_{j})\phi ^{-1} \bigl(P^{-1}(d _{j}-c_{j}) \bigr). $$
(2.11)
Furthermore,
$$\begin{aligned} \psi \bigl(-u' \bigr) \leq &- \int _{C_{-}}u'(t)\,dt=-\sum _{j\in I_{-}} \int _{c_{j}}^{d_{j}}u'(t)\,dt \\ \leq &\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(d_{j}-c_{j}) \bigr)\mid j\in I_{-} \bigr\} \sum_{j\in I_{+}}(d_{j}-c_{j}) \\ \leq &\mu (C_{-})\sup \bigl\{ \phi ^{-1} \bigl(P^{-1}(d_{j}-c_{j}) \bigr)\mid j \in I_{-} \bigr\} . \end{aligned}$$
Therefore, (2.8) is satisfied.
The result follows now from (2.4), (2.5) and (2.8). □
Let us consider the homotopy problem
$$ \bigl(\phi \bigl(u'(t) \bigr) \bigr)'=\lambda (Fu) (t), \quad \lambda \in [0,1], $$
(2.12)
depending on the parameter λ.
The next lemma gives prior bounds for solutions of (2.12) and (1.2).
Lemma 2.8
Suppose thatuis a solution of (2.12) for any
\(\lambda \in [0,1]\)and satisfies the boundary condition (1.2) with
\(A=0\). Then the following conclusions are fulfilled:
$$\begin{aligned}& \Vert u \Vert _{\infty }\leq B, \end{aligned}$$
(2.13)
$$\begin{aligned}& \Vert u \Vert _{C^{1}}\leq B+a. \end{aligned}$$
(2.14)
Proof
From \(\omega (u)=A=0\) and Lemma 2.3, there exists a \(\delta \in [0,T]\) such that \(u(\delta )=0\). Thus
$$ \max \bigl\{ u(t) \mid t\in [0,T] \bigr\} \geq 0, $$
this together with (2.1) shows that we obtain (2.13).
Taking into account \(\phi :(-a,a)\) and (2.13), we deduce that
$$ \Vert u \Vert _{C^{1}}= \Vert u \Vert _{\infty }+ \bigl\Vert u' \bigr\Vert _{\infty }< B+a. $$
□
We now state the following important lemma.
Lemma 2.9
LetBbe a positive constant, \(\omega \in \mathcal{A}\)andψbe defined in (2.2). Set
$$ \varOmega = \bigl\{ (u,\alpha ,\beta )\mid (u,\alpha ,\beta )\in Y\times \mathbb{R}^{2}, \Vert u \Vert _{C^{1}}< \rho , \bigl\Vert u' \bigr\Vert _{\infty }< a, \vert \alpha \vert < \rho , \vert \beta \vert < \phi (a) \bigr\} , $$
where
\(\rho =B+a\)and
\(\rho < aT\).
Define
\(\varPhi _{i}: \overline{\varOmega }\rightarrow Y\times \mathbb{R} ^{2}\) (\(i=1,2\)),
$$\begin{aligned}& \varPhi _{1}(u,\alpha ,\beta )= \bigl(\alpha +\phi ^{-1}( \beta )t,\alpha + \omega (u), \beta +\psi \bigl(u' \bigr)-B \bigr), \end{aligned}$$
(2.15)
$$\begin{aligned}& \varPhi _{2}(u,\alpha ,\beta )= \bigl(\alpha +\phi ^{-1}( \beta )t,\alpha + \omega (u), \beta +\psi \bigl(-u' \bigr)-B \bigr). \end{aligned}$$
(2.16)
Then
$$ D(I-\varPhi _{i},\varOmega ,0)\neq 0,\quad i=1,2, $$
(2.17)
whereD, Idenote the Leray–Schauer degree and the identity operator on
\(Y\times \mathbb{R}^{2}\), respectively.
Proof
Obviously, Ω is a bounded open subset of the Banach space \(Y\times \mathbb{R}^{2}\) with usual norm, and it is symmetric with respect to \(\theta \in \varOmega \).
Define \(G_{i}:[0,1]\times \varOmega \rightarrow Y\times \mathbb{R}^{2}\) (\(i=1,2\)),
$$\begin{aligned}& \begin{aligned} G_{1}(\lambda ,u,\alpha ,\beta )={} & \bigl(\alpha + \bigl(\phi ^{-1}(\beta )-(1- \lambda ) \phi ^{-1}(-\beta ) \bigr)t,\alpha +\omega (u)-(1-\lambda )\omega (-u), \\ &\beta +\psi \bigl(u' \bigr)-\psi \bigl((\lambda -1)u' \bigr)-\lambda B \bigr), \end{aligned} \\& \begin{aligned} G_{2}(\lambda ,u,\alpha ,\beta )={} & \bigl(\alpha + \bigl(\phi ^{-1}(\beta )-(1- \lambda ) \phi ^{-1}(-\beta ) \bigr)t,\alpha +\omega (u)-(1-\lambda )\omega (-u), \\ &\beta +\psi \bigl(-u' \bigr)-\psi \bigl((1-\lambda )u' \bigr)-\lambda B \bigr). \end{aligned} \end{aligned}$$
For all \((u,\alpha ,\beta )\in \overline{\varOmega }\), it is clear that \(G_{i}(1,u,\alpha ,\beta )=\varPhi _{i}(u,\alpha ,\beta )\) (\(i=1,2\)). Hence to prove \(D(I-\varPhi _{i},\varOmega ,0)\neq 0\), we only need to prove the following hypotheses holding by the Borsuk theorem [13, Theorem 8.3].
- (1)
\(G_{i}(0,\cdot ,\cdot ,\cdot )\) is an odd operator on Ω̅, that is,
$$ G_{i}(0,-u,-\alpha ,-\beta )=-G_{i}(0,u,\alpha ,\beta ) \quad (i=1,2), (u,\alpha ,\beta )\in \overline{\varOmega }; $$
(2.18)
- (2)
\(G_{i}\) is a completely continuous operator;
- (3)
\(G_{i}(\lambda ,u,\alpha ,\beta )\neq (u,\alpha ,\beta )\) for \((\lambda ,u,\alpha ,\beta )\in [0,1]\times \partial \varOmega \).
In fact, we take \((u,\alpha ,\beta )\in \overline{\varOmega }\), for \(i=1\),
$$ \begin{aligned} G_{1}(0,-u,-\alpha ,-\beta ) &= \bigl(- \alpha + \bigl(\phi ^{-1}(-\beta )- \phi ^{-1}(\beta ) \bigr)t,-\alpha +\omega (-u)-\omega (u), \\ &\quad {-}\beta +\psi \bigl(-u' \bigr)- \psi \bigl(u' \bigr) \bigr) \\ &=-G_{1}(0,u,\alpha ,\beta ). \end{aligned} $$
Analogously \(G_{2}(0,-u,-\alpha ,-\beta )=-G_{2}(0,u,\alpha ,\beta )\). So (1) is asserted.
Next we show that (2) holds.
Let \(\{(\lambda _{n},u_{n},\alpha _{n},\beta _{n})\}\subset [0,1]\times \overline{\varOmega }\) be a sequence. Then, for each \(n\in \mathbb{Z} ^{+}\) and by the fact that \(t\in [0,T]\), \(0\leq \lambda _{n}\leq 1\), \(\|u _{n}\|_{C^{1}}<\rho \), \(|\alpha _{n}|\leq \rho \), \(|\beta _{n}|<\phi (a)\); meanwhile, \(\{\omega (u_{n})\}\), \(\{\omega (-u_{n})\}\), \(\{\psi (u_{n})\}\) and \(\{\psi (-u_{n})\}\) are bounded. By the Arzelà–Ascoli theorem, it is not difficult to verify they are relatively compact. Then \(G_{i}(\lambda ,u,\alpha ,\beta )\) is convergent in \(Y\times \mathbb{R}^{2}\). It follows from the continuity of \(\phi ^{-1}\), ω and ψ that \(G_{i}\) (\(i=1,2\)) is continuous. So \(G_{i}\) (\(i=1,2\)) are completely continuous.
Finally, we prove that (3) is valid. Assume on the contrary that
$$ G_{i}(\lambda _{0},u_{0},\alpha _{0},\beta _{0})=(u_{0},\alpha _{0},\beta _{0}) $$
(2.19)
for some \((\lambda _{0},u_{0},\alpha _{0},\beta _{0})\in [0,1]\times \partial \varOmega \). Then
$$\begin{aligned}& \alpha _{0}+ \bigl(\phi ^{-1}(\beta _{0})-(1- \lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)t=u _{0}(t), \end{aligned}$$
(2.20)
$$\begin{aligned}& \omega (u_{0})-(1-\lambda _{0})\omega (-u_{0})=0, \end{aligned}$$
(2.21)
$$\begin{aligned}& \psi \bigl(u'_{0} \bigr)-\psi \bigl((\lambda _{0}-1)u'_{0} \bigr)=\lambda _{0} B. \end{aligned}$$
(2.22)
By Lemma 2.3 (take \(u=u_{0}\), \(k=1-\lambda _{0}\)) and (2.21), there exist \(\gamma \in [0,T]\) and, consequently, \(u_{0}(\gamma )=0\). Together with (2.20) this shows that we obtain
$$ \alpha _{0}=- \bigl(\phi ^{-1}(\beta _{0})-(1- \lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr) \gamma $$
(2.23)
and
$$ u_{0}(t)= \bigl(\phi ^{-1}(\beta _{0})-(1- \lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr) (t- \gamma ). $$
(2.24)
The rest of the proof is divided into three cases.
Case 1. If \(\beta _{0} =0\), it follows from (2.23), (2.24) that \(\alpha _{0} =0\), \(u_{0}=0\), then
$$ (0,0,0)=(u_{0},\alpha _{0},\beta _{0})\in \partial \varOmega , $$
which is a contradiction.
Case 2. If \(\beta _{0} >0\), one deduces from \(\phi ^{-1}(\beta _{0})-(1- \lambda _{0})\phi ^{-1}(-\beta _{0})>0\) and the definition of ψ in (2.2) that
$$ \psi \bigl(u'_{0} \bigr)-\psi \bigl((\lambda _{0}-1)u'_{0} \bigr)= \bigl(\phi ^{-1}(\beta _{0})- (1- \lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T. $$
Combining this with (2.22), we have
$$ \bigl(\phi ^{-1}(\beta _{0})- (1-\lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T=\lambda _{0}B $$
(2.25)
and
$$ \phi ^{-1}(\beta _{0})\leq \frac{\lambda _{0}\rho }{T}, \quad \text{if } -(1-\lambda _{0})\phi ^{-1}(-\beta _{0})\geq 0. $$
Hence, \(\beta _{0}\leq \phi (\frac{\lambda _{0}\rho }{T})< \phi (a)\).
On the other hand, according to (2.23)–(2.25), for each \(t\in [0,T]\), we conclude that
$$ \begin{aligned} & \bigl\vert u_{0}(t) \bigr\vert \leq \frac{\lambda _{0}B}{T} \vert t-\gamma \vert \leq B, \\ & \bigl\vert u'_{0}(t) \bigr\vert = \phi ^{-1}(\beta _{0})-(1-\lambda _{0})\phi ^{-1}(-\beta _{0})\leq \frac{ \lambda _{0}B}{T}\leq \frac{\rho }{T}< a, \\ & \vert \alpha _{0} \vert = \bigl\vert u_{0}(0) \bigr\vert < \Vert u _{0} \Vert _{\infty }< \rho , \qquad \Vert u_{0} \Vert _{C^{1}}< B+a=\rho . \end{aligned} $$
Thus \((u_{0},\alpha _{0},\beta _{0})\notin \partial \varOmega \), a contradiction.
Case 3. If \(\beta _{0} <0\), it follows that \(\phi (\beta '_{0})- \phi ((\lambda _{0}-1)\beta '_{0})<0\), and by the definition of ψ in (2.2), we obtain
$$ \begin{aligned} \psi \bigl(u'_{0} \bigr)- \psi \bigl((\lambda _{0}-1)u'_{0} \bigr) &=0-(\lambda _{0}-1) \bigl(\phi ^{-1}(\beta _{0})- (1-\lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T \\ &=(1- \lambda _{0}) \bigl(\phi ^{-1}(\beta _{0})- (1-\lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T. \end{aligned} $$
Combining this with (2.22), we deduce that
$$ (1-\lambda _{0}) \bigl(\phi ^{-1}(\beta _{0})- (1-\lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T=\lambda _{0}B. $$
(2.26)
If \(\lambda _{0}=0\), then (2.26) implies \(\phi ^{-1}(\beta _{0})- \phi ^{-1}(-\beta _{0})=0\), which contradicts \(\phi (\beta '_{0})-\phi (- \beta '_{0})<0\).
If \(\lambda _{0}=1\), then \(\lambda _{0}B=0\), i.e. \(B=0\), which is impossible.
If \(\lambda _{0}\in (0,1)\), then
$$ (1-\lambda _{0}) \bigl(\phi ^{-1}(\beta _{0})- (1-\lambda _{0})\phi ^{-1}(-\beta _{0}) \bigr)T< 0, \quad \text{also } \lambda _{0}B>0. $$
This is a contradiction. The proof is completed. □