Throughout this notebook, latin letters replace greek ones.

To begin with, we define ϕ as p, that is,

p=m+u, where u:=(OverBar[u])=2, and m = μ.

---------FIRST STEP, SETTING UP THE SYSTEM OF EQUATIONS-----------------------------------

We normalize here t1=0, so in terms of T1=exp(-t_1)

T1=1;

More generally, we write Ti for exp(-t_i)

We let di :=exp(δ_i). Eqn. (4) of the proof in appendix becomes

X1=(m/(m+1))((1-d3 (T3^m))/(1-(T3^m)));

where X1 :=exp(-ℓ_1)/γ. Equation(3) becomes

X2= (1+C1 T1),

where X2=(1+C1 T2)/d2

Let us write condition (5) of the file (the soc at t3) as the root of:

X3=p-1-(m+1)X1(1/d3)-(1/(1+D2  T3)).

Next we write the two conditions of (6)-(7) (the soc for player 2) as solutions to A1=0, B1=0, as defined next. Note that e^{-ℓ_2}/gamma=X1 X2/(1+C1 T2).

A1=(1+C1 T2)((p-1)/p)-X1 X2-((1-((T3^((p-1)))/(T2^((p-1)))))/(p(p-1)));

B1=(1+C1 T2)((p-1)/p)-X1 X2-((1-((T3^((p-1)))/(T2^((p-1)))))/(p(p-1)))-((T3^((p-1)))/(T2^((p-1))))(1+C1 T3)(((1+C1 T2)/(1+C1 T3))(((T3^(2-p))/(T2^(2-p)))(1-((X1 X2)/(1+C1 T2)))+((X1 X2)/(1+C1 T2))(1/m)((T3^(2-u))/(T2^(2-u)))(1-((T3^(-m))/(T2^(-m)))))-(1/p));

Next we write the first foc at t1 (combined with t3), namely (1) of the proof. The last term exploits equation (2)  ("learning at ℓ_3")

A2=FullSimplify[-(1-(m/(m+1))((1-d3 (T3^m))/(1-(T3^m))))+(((T2^(p-1))-1)/(1-p))+(1/(1+D2 T2))(-(((T2^(-1+p))-(T3^(-1+p)))/(1-p))+((D2 ((T2^p)-(T3^p)))/p))+(T3^(p-1))((1+D2 T3)/(1+D2 T2))(1-(m/(m+1))((1-d3 (T3^m))/(1-(T3^m))) (1/d3))]

Finally we write equation (2) ("Learning at ℓ_3") as the root of the following expression, called A3:

A3=((C2+(1/T2))/(D2+(1/T2)))(1+D2 T3)(1/d3)-(T3^(2-u))(1+C2 T1);

---------SECOND STEP: SIMPLIFYING/RELAXING THE SET OF CONSTRAINTS-----------------------------------

We first solve for A2 and A3.

Simplify[Solve[{A2==0,A3==0},{D2,d3}]]

We note that the first solution is not admissible, given D2>0. Hence we focus on the second.

We define the remaining constraints inserting the values of D3 and d3: Simplify[{A1≥0,B1≥0,X3≥0,((p-1)/p)-(X1/d3)≥0,D2-C2≥0,C1-C2≥0, 1>T2>T3>0,1>d3>0}/.{D2->(C2 m (2+m) (-1+T2) (T3^m) (-1+T3+C2 (-T2+T3)))/(C2 m (2+m) (T2^2) (-1+C2 (-1+T3)) (T3^m)-(1+C2) (T2^(2+m)) (-1+(T3^m))-C2 (1+C2) (T2^(3+m)) (-1+(T3^m))+(T3^(1+m)) (-2 m (-1+T3+C2 T3)-(m^2) (-1+T3+C2 T3)+(1+C2) T3 (-1+(T3^m)))+T2 (T3^m) (C2 (1+C2) (T3^2) (-1+(T3^m))-2 m (1-T3+(C2^2) (-1+T3) T3+C2 (1-4 T3+(T3^2)))-(m^2) (1-T3+(C2^2) (-1+T3) T3+C2 (1-4 T3+(T3^2))))),d3->(C2 m (2+m) (T2^2) (T3^m)+(T2^(2+m)) (-1+(T3^m))+C2 (T2^(3+m)) (-1+(T3^m))+(T3^(1+m)) (2 m (-1+T3+C2 T3)+(m^2) (-1+T3+C2 T3)-T3 (-1+(T3^m)))-T2 (T3^m) (2 m (-1+T3+2 C2 T3)+(m^2) (-1+T3+2 C2 T3)+C2 (T3^2) (-1+(T3^m))))/(C2 m (2+m) (T2^2) (T3^m)-m (2+m) T2 (T3^m) (-1+T3+2 C2 T3)+(1+C2) (T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (2 m (-1+T3+C2 T3)+(m^2) (-1+T3+C2 T3)-(1+C2) T3 (-1+(T3^m))))}]

This gives us Z1:

Z1[m_]:=Z1[m]={((- ((1+C2) (T2^(2+m))+(T3^(1+m)) (-(2+m) (1+C2 T2)+(1+C2) (1+m) T3)))/( (-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2))))))≤0,(1+C2) (T2^(2+m)) (1+C2 T2)+(1+C2) (1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)-(1+C2) (1+C2 T2) ((T2^(2+m))+(T3^2)))≥0,C2>0,1>T2>T3>0,1>((-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2))))/(C2 m (2+m) (T2^2) (T3^m)-m (2+m) T2 (T3^m) (-1+T3+2 C2 T3)+(1+C2) (T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) ((1+C2) T3-(1+C2) (T3^(1+m))+m (2+m) (-1+T3+C2 T3))))>0}

Consider the last constraint (the term between 0 and 1) and note that

FullSimplify[1-((-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2))))/(C2 m (2+m) (T2^2) (T3^m)-m (2+m) T2 (T3^m) (-1+T3+2 C2 T3)+(1+C2) (T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) ((1+C2) T3-(1+C2) (T3^(1+m))+m (2+m) (-1+T3+C2 T3))))]

hence we can rewrite our set of constraints as:

Z2[m_]:=Z2[m]={((- ((1+C2) (T2^(2+m))+(T3^(1+m)) (-(2+m) (1+C2 T2)+(1+C2) (1+m) T3)))/( (-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2))))))≤0,(1+C2) (T2^(2+m)) (1+C2 T2)+(1+C2) (1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)-(1+C2) (1+C2 T2) ((T2^(2+m))+(T3^2)))≥0,C2>0,1>T2>T3>0,1>((C2 (-1+T2) (-1+(T3^m)) ((T2^(2+m))-(T3^(2+m))))/(-C2 m (2+m) (T2^2) (T3^m)+m (2+m) T2 (T3^m) (-1+T3+2 C2 T3)-(1+C2) (T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)-(1+C2) (((1+m))^2) T3+(1+C2) (T3^(1+m)))))>0}

Let us relax the last constraint by dropping "1>..." and define Z3 accordingly:

Z3[m_]:=Z3[m]={-(((1+C2) (T2^(2+m))+(T3^(1+m)) ((-2-m) (1+C2 T2)+(1+C2) (1+m) T3))/(-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2)))))≤0,(1+C2) (T2^(2+m)) (1+C2 T2)+(1+C2) (1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)-(1+C2) (1+C2 T2) ((T2^(2+m))+(T3^2)))≥0,C2>0,1>T2>T3>0,((C2 (-1+T2) (-1+(T3^m)) ((T2^(2+m))-(T3^(2+m))))/(-C2 m (2+m) (T2^2) (T3^m)+m (2+m) T2 (T3^m) (-1+T3+2 C2 T3)-(1+C2) (T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)-(1+C2) (((1+m))^2) T3+(1+C2) (T3^(1+m)))))>0}

Given the sign of the numerator of the last expression, we might alternatively work with Z4, defined as:

Z4[m_]:=Z4[m]={-(((1+C2) (T2^(2+m))+(T3^(1+m)) ((-2-m) (1+C2 T2)+(1+C2) (1+m) T3))/(-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2)))))≤0,(1+C2) (T2^(2+m)) (1+C2 T2)+(1+C2) (1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)-(1+C2) (1+C2 T2) ((T2^(2+m))+(T3^2)))≥0,C2>0,1>T2>T3>0,(T3^m) (m (2+m) T2 (-1+T3+2 C2 T3)+T3 (m (2+m)-(1+C2) (((1+m))^2) T3+(1+C2) (T3^(1+m))))-(T2^2) (C2 m (2+m) (T3^m)+(1+C2) (T2^m) (-1+(T3^m)))>0};

Let us manipulate these constraints as follows. First, the first constraint states that a1/a2≥0, where

a1=(1+C2) (T2^(2+m))+(T3^(1+m)) (-(2+m) (1+C2 T2)+(1+C2) (1+m) T3);

a2=-(T2^(2+m)) (1+C2 T2)-(1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2-T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2-T3) (1+C2 T2-(1+C2) T3)+(1+C2 T2) ((T2^(2+m))+(T3^2)));

while the last states that a4>0, where

a4=(T3^m) (m (2+m) T2 (-1+T3+2 C2 T3)+T3 (m (2+m)-(1+C2) (((1+m))^2) T3+(1+C2) (T3^(1+m))))-(T2^2) (C2 m (2+m) (T3^m)+(1+C2) (T2^m) (-1+(T3^m)));

We note that

FullSimplify[a2+a4]

which is the numerator of the last expression of Z3, which we know is less than 1. Hence, it must be that a1<0. We write:

Z6[m_]:=Z6[m]={((1+C2) (T2^(2+m))+(T3^(1+m)) ((-2-m) (1+C2 T2)+(1+C2) (1+m) T3))≤0,(1+C2) (T2^(2+m)) (1+C2 T2)+(1+C2) (1+C2 T2) (T3^(2+2 m))+(T3^m) (2 m (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)+(m^2) (T2 (-1+C2 (-1+T3))+T3) (1+C2 T2-(1+C2) T3)-(1+C2) (1+C2 T2) ((T2^(2+m))+(T3^2)))≥0,C2>0,1>T2>T3>0,(T3^m) (m (2+m) T2 (-1+T3+2 C2 T3)+T3 (m (2+m)-(1+C2) (((1+m))^2) T3+(1+C2) (T3^(1+m))))-(T2^2) (C2 m (2+m) (T3^m)+(1+C2) (T2^m) (-1+(T3^m)))>0};

We now do a simple change of variables, defining x by :

Solve[1+C2 T2==(1+C2)x,C2]

which must be less than 1, by definition (T2<1).

We have that

FullSimplify[T2-((1+C2 T2)/(1+C2))]

<0 , hence  T2-x<0

Rewriting the conditions using x gives us:

Z7[m_]:=Z7[m]={(T2^(2+m))+(T3^(1+m)) ((1+m) T3-(2+m) x)<0,(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x+2 m (T3^m) (T3-x) (T2-T3 x)+(m^2) (T3^m) (T3-x) (T2-T3 x)>0,1>T2>T3>0,2 m (T2-T3) (T3^m) (T3-x)+(m^2) (T2-T3) (T3^m) (T3-x)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x>0,1>x};

-----------------------------------------------------------------------------LAST STEP: THERE IS NO SOLUTION TO Z7--------------------------------------------------------------------

Claim:The system defined by:

Z7[m_]:=Z7[m]={(T2^(2+m))+(T3^(1+m)) ((1+m) T3-(2+m) x)<0,(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x+2 m (T3^m) (T3-x) (T2-T3 x)+(m^2) (T3^m) (T3-x) (T2-T3 x)>0,1>T2>T3>0,2 m (T2-T3) (T3^m) (T3-x)+(m^2) (T2-T3) (T3^m) (T3-x)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x>0,1>x};

admits no solution {x,T2,T3,m},where m>0, and x≤1. We will add the constraint x>0 which we know must be satisfied (in fact, we know x>T2).

The logic is as follows: Z7 is a linear system in x, except for the second term, which is convex in x, and is ≥0 inequality. Hence if it can be satisfied, it can be satisfied (relaxing the strict constraints to weak ones) with equality in x, for at least one of the constraints that is linear in x (if not, simply shift x in whichever direction increases the convex term). Hence, we may consider the following three cases:

--------------------------------1St CASE: THE CONSTRAINT x≤1 BINDS-------------------------------------------

If so, let us simplify:

Z7[m]/.x->1

Simplify[{(T2^(2+m))+(T3^(1+m)) (-2-m+(1+m) T3)<0,2 m (T2-T3) (-1+T3) (T3^m)+(m^2) (T2-T3) (-1+T3) (T3^m)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m)))>0,1>T2>T3>0,2 m (T2-T3) (-1+T3) (T3^m)+(m^2) (T2-T3) (-1+T3) (T3^m)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m)))>0}]

Let us focus on the non-trivial constraints, defining ya:

ya[m_]:=ya[m]={(T2^(2+m))+(T3^(1+m)) (-2+m (-1+T3)+T3)<0,(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))>(T2^(2+m)) (-1+(T3^m)),1>T2>T3>0};

We note that the second constraint satisfies:

FullSimplify[D[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m)),T2]]

so the second term is convex in T2, and it's enough to check for the two extreme cases, T2=T3, T2 = upper bound from first term. Yet at T2=T3:

FullSimplify[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m))/.T2->T3]

so this does not work. Let us condition the other extreme, solving the equation for T2:

FullSimplify[Solve[(T2^(2+m))+(T3^(1+m)) (-2-m+(1+m) T3)==0,T2],Assumptions->m>0&&T3>0]

and plugging in:

FullSimplify[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m))/.T2->(T3^((1+m)/(2+m)))(((-(-2+m (-1+T3)+T3)))^(1/(2+m)))]

One might immediately be reassured by plotting for (m,T3) and a formal proof follows:

Plot3D[Sign[-((((T3^((1+m)/(2+m))) (((2+m-(1+m) T3))^(1/(2+m)))))^(2+m)) (-1+(T3^m))+(T3^m) (m (2+m) (-1+T3) (T3^((1+m)/(2+m))) (((2+m-(1+m) T3))^(1/(2+m)))+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))],{m,.01,5},{T3,0,1}]

Formally, we note that, simplifying the expression (we know the signs of the terms elevated to powers), we have

FullSimplify[-((T3^(1+m)) (2+m-(1+m) T3)) (-1+(T3^m))+(T3^m) (m (2+m) (-1+T3) (T3^((1+m)/(2+m))) (((2+m-(1+m) T3))^(1/(2+m)))+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))]

Simplifying the inner term,

FullSimplify[-(-(1+m) (T3^(2+m))+(T3^((1+m)(2+m)))+m (T3^(1+m)) (2+m-(1+m) T3))]

which, as a function of T3:

FullSimplify[D[m (2+m) (-1+T3)+T3-(T3^(((1+m))^2)),{T3,2}]]

is concave

Hence, we check at both end points. At T3=1, we have that the derivative

FullSimplify[D[m (2+m) (-1+T3)+T3-(T3^(((1+m))^2)),{T3,1}]/.T3->1,Assumptions->m>0]

is zero, so the function is flat at 1, so the maximum is achieved there, yet

FullSimplify[m (2+m) (-1+T3)+T3-(T3^(((1+m))^2))/.T3->1]

so the expression is always negative.

-------------------------------2ND CASE: THE   CONSTRAINT (T2^(2+m))+(T3^(1+m)) ((1+m) T3-(2+m) x)==0 BINDS--------------------------------------------

This means that:

Solve[(T2^(2+m))+(T3^(1+m)) ((1+m) T3-(2+m) x)== 0,x]

We plug nack and simplify:

FullSimplify[Z7[m]/.{x->(((T3^(-1-m)) ((T2^(2+m))+(T3^(2+m))+m (T3^(2+m))))/(2+m))}]

we simplify these constraints by getting rid of the factors whose sign is known:

Simplify[{- (m (2+m) T2 (T3^m)+(T2^(2+m)) (-1-m+(T3^m))+(1+m) (T3^(2+m)) (-1-m+(T3^m)))>0,1>T2>T3>0,-(m (2+m) T2 (T3^m)+(1+m) (T3^(2+2 m))+(T2^(2+m)) (-1+(T3^m))-(T3^(1+m)) (T3+m (2+m+T3)))>0,T3+m T3+(T2^(2+m)) (T3^(-1-m))<2+m}]

We focus on the first constraint

yb[m_]:=yb[m]={- (m (2+m) T2 (T3^m)+(T2^(2+m)) (-1-m+(T3^m))+(1+m) (T3^(2+m)) (-1-m+(T3^m)))>0,1>T2>T3>0,(T2^(2+m))+(T3+m T3-(2+m)) (T3^(1+m))<0};

We note that this constraint is convex in  T2:

Simplify[D[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m)),{T2,2}]]

So again, we have a constraint on T2, between T3 and the upper bound from the first equation (which is the same as for case ya):

We note that at the lower end,

FullSimplify[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m))/.T2->T3]

while at the upper end, we get back from the formula from the first case, namely:

FullSimplify[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m))/.T2->(T3^((1+m)/(2+m)))(((-(-2+m (-1+T3)+T3)))^(1/(2+m)))]

which is the same as before, see above:

-((((T3^((1+m)/(2+m))) (((2+m-(1+m) T3))^(1/(2+m)))))^(2+m)) (-1+(T3^m))+(T3^m) (m (2+m) (-1+T3) (T3^((1+m)/(2+m))) (((2+m-(1+m) T3))^(1/(2+m)))+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))

hence we get again an impossibility.

------------------------------THIRD CASE:THE CONSTRAINT 2 m (T2-T3) (T3^m) (T3-x)+(m^2) (T2-T3) (T3^m) (T3-x)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x==0  BINDS--------------------------------------------------------------------------------------

Recall that:

Z7[m]

We now assume that it the penultimate equation that binds:

FullSimplify[Solve[2 m (T2-T3) (T3^m) (T3-x)+(m^2) (T2-T3) (T3^m) (T3-x)+(-1+(T3^m)) (-(T2^(2+m))+(T3^(2+m))) x==0,x]]

We plug back  in Z7:

FullSimplify[Z7[m]/.x->-((m (2+m) (T2-T3) (T3^(1+m)))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))]

Again, we ignore factors whose sign is known, and define the remaining constraints as yc, where:

yc[m_]:=yc[m]={(T2^(2+m))+(T3^(2+m)) (1+m+((m (((2+m))^2) (T2-T3) (T3^m))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m))))))<0,(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))>(T2^(2+m)) (-1+(T3^m)),1>T2>T3>0,-((m (2+m) (T2-T3) (T3^(1+m)))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))<1};

Gathering the terms in the first constraint gives:

FullSimplify[Together[(T2^(2+m))+(T3^(2+m)) (1+m+((m (((2+m))^2) (T2-T3) (T3^m))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m))))))]]

Subtracting one yields

FullSimplify[(( (m (2+m) T2 (T3^m)+(1+m) (T3^(2+2 m))+(T2^(2+m)) (-1+(T3^m))-(T3^(1+m)) (T3+m (2+m+T3))))/(m (2+m) T2 (T3^m)+(T2^(2+m)) (-1+(T3^m))-(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))-1]

Hence the constraints yc are equivalent to:

{(((2+m) (T3^(2+m)) (-1+(T3^m)))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))>1,(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))>(T2^(2+m)) (-1+(T3^m)),1>T2>T3>0,-((m (2+m) (T2-T3) (T3^(1+m)))/(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))<1};

so in particular we must have  from the first constraint

-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))<0 for the ratio to be positive.

Hence, simplifying:

FullSimplify[{(2+m) (T3^(2+m)) (-1+(T3^m))<-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m))),(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))>(T2^(2+m)) (-1+(T3^m)),1>T2>T3>0,-m (2+m) (T2-T3) (T3^(1+m))>-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m))),-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))<0}]

Note that the sum of 2nd term and 3rd term  cancel:

FullSimplify[(T3^m) (m (2+m) T2 (-1+T3)+T3 (-m (2+m) (-1+T3)-T3+(T3^(1+m))))-(T2^(2+m)) (-1+(T3^m))+(-m (2+m) (T2-T3) (T3^(1+m))-(-m (2+m) T2 (T3^m)-(T2^(2+m)) (-1+(T3^m))+(T3^(1+m)) (m (2+m)+T3 (-1+(T3^m)))))]

hence it is impossible that both be positive. We are done.

----------------------------------------------------LAST CASE: x binds at 0----------------------------------------

yc=FullSimplify[Z7[m]/.x->0]

and the first constraint is impossible to satisfy.