Without loss of generality we may assume 𝑓(1) =1,
since otherwise we may use 𝑔(𝑥) =𝑓(1)−1𝑓(𝑥)
First we will show that if such a 𝜑 exists it is necessarily unique.
Let 𝜑,𝜓 :[0,1] →ℝ with 𝜑(0) =𝜓(0) =0 and ex𝜑 =ex𝜓 =𝑓ex.
Then
ex𝜑ex𝜓(𝑡)=ex(𝜑(𝑡)−𝜓(𝑡))=1for all 𝑡 ∈[0,1],
which may be the case iff 𝜑(𝑡) −𝜓(𝑡) ∈ℤ for all 𝑡 ∈[0,1].
Since 𝜑 and 𝜓 are continuous so is 𝜑 −𝜓,
and thus 𝜑 −𝜓 is a constant map.
Thus (𝜑 −𝜓)(𝑡) =(𝜑 −𝜓)(0) =0 for all 𝑡 ∈[0,1],
i.e. 𝜑 =𝜓.
Since 𝑓ex :[0,1] →𝕊1 is continuous it is uniformly continuous by the Heine-Cantor theorem,
we can divide [0,1] by 0 =𝑡0 <𝑡1 <⋯ <𝑡𝑘 =1 with finite 𝑘 so that
𝑡∈[𝑡𝑗,𝑡𝑗+1]⟹|𝑓ex(𝑡)−𝑓ex(𝑡𝑗)|<2for all integers 0 ≤𝑗 ≤𝑘 −1.
We write 𝑗(𝑡) to denote the value of 𝑗 corresponding to some 𝑡.
whence it follows that 𝑓ex(𝑡) and 𝑓ex(𝑡𝑗(𝑡)) are not antipodes, namely
𝑓ex(𝑡)𝑓ex(𝑡𝑗(𝑡))≠−1and therefore the Main branch of the complex logarithm Ln(𝑓ex(𝑡)/𝑓ex(𝑡𝑗(𝑡))) is well-defined.
We define 𝜑 as follows
𝜑(𝑡)=12𝜋𝑖𝑗(𝑡)∑𝑛=1Ln𝑓ex(𝑡𝑛)𝑓ex(𝑡𝑛−1)+Ln𝑓ex(𝑡)𝑓ex(𝑡𝑗(𝑡))which is continuous by properties of the Main branch of the complex logarithm.
Additionally, 𝜑(0) =0 and clearly ex𝜑 =𝑓ex.
All that's left to show is that 𝜑(1) ∈ℤ.
This is true since by definition 𝑓(1)ex𝜑(1) =𝑓ex(1) =𝑓(1) and hence ex𝜑(1) =1 ⟹ 𝜑(1) ∈ℤ.