By [Ber04, Theorem 5.1.1] we know that $|f(x)|\neq 0$ and that $\varphi$ is piecewise affine linear on $\Delta$. By [Ber99, Theorem 5.2] we have $\varphi\leq\varphi\circ p_{\mathfrak{X}}$. Assume there is a face $\tau$ of $\Delta$ on which $\varphi$ is not convex, i.e. there are $x,y\in\tau$ and $t\in(0,1)$ such that

$$\delta:=\varphi(tx+(1-t)y)-t\varphi(x)-(1-t)\varphi(y)>0.$$

By base change we can assume that $K$ is algebraically closed and then by density of the value group $\Gamma$ and continuity of $\varphi$ that the coordinates of $x$ and $y$ are in $\Gamma$. Choose a $\Gamma$-rational polytopal subdivision of $\Delta$ which only has $x$ and $y$ as additional vertices. By Construction 2.6 we get an admissible formal model $\mathfrak{X}^{\prime\prime}$ of $\mathfrak{X}^{\textup{an}}$ dominating $\mathfrak{X}$. Choose an affine open $U\subseteq\tilde{\mathfrak{X}}^{\prime\prime}$ which contains $\red(tx+(1-t)y)$. By the stratum face correspondence (Proposition 2.8 and Corollary 2.9) the vertices $x$ and $y$ correspond to irreducible components of $\tilde{\mathfrak{X}}^{\prime\prime}$. By taking out all other irreducible components we may assume that $U$ intersects only those corresponding to $x$ and $y$. Then $V:=\red^{-1}(U)$ is a strictly $K$-affinoid domain by [Bos77, Theorem 3.1]. By [Ber99, Proposition 1.4] its canonical reduction has two irreducible components, namely those corresponding to $x$ and $y$. Hence the Shilov boundary of $V$ is the set $\{x,y\}$ by [Ber90, Proposition 2.4.4] and we get $|f(tx+(1-t)y)|\leq\max\left\{|f(x)|,|f(y)|\right\}$. Since $x\neq y$, by restricting to a building block $\mathfrak{U}$, we can find a coordinate function $g\in\mathcal{O}(\mathfrak{U}^{\textup{an}})^{\times}$ such that $|g(x)|\neq|g(y)|$. Then we can find $N\in\mathbb{N}_{>0}$ and $m\in\mathbb{Z}$ such that

$$\Big|\log|f^{N}g^{m}(x)|-\log|f^{N}g^{m}(y)|\Big|=\Big|N(\varphi(x)-\varphi(y))+m(\log|g(x)|-\log|g(y)|)\Big|<N\delta.$$

Since $\log|g^{m}|$ is affine linear on $\tau$ we get

$$\log|f^{N}g^{m}(tx+(1-t)y)|-t\log|f^{N}g^{m}(x)|-(1-t)\log|f^{N}g^{m}(y)|=N\delta.$$

Hence by replacing $f$ with $f^{N}g^{m}$ and $\delta$ by $N\delta$ we can assume

$$\delta:=\varphi(tx+(1-t)y)-t\varphi(x)-(1-t)\varphi(y)>0.$$

and

$$|\varphi(x)-\varphi(y)|<\delta.$$

Then

$$\varphi(tx+(1-t)y)=\delta+t\varphi(x)+(1-t)\varphi(y)>t\varphi(x)+(1-t)\varphi(y)+|\varphi(x)-\varphi(y)|.$$

Now on the one hand we have

$$t\varphi(x)+(1-t)\varphi(y)+|\varphi(x)-\varphi(y)|\geq t\varphi(x)+(1-t)\varphi(y)+t(\varphi(y)-\varphi(x))=\varphi(y)$$

while on the other hand

$$t\varphi(x)+(1-t)\varphi(y)+|\varphi(x)-\varphi(y)|\geq\varphi(x)+t(\varphi(x)-\varphi(y)).$$

Together we get

$$\varphi(tx+(1-t)y)>\max\left\{\varphi(x),\varphi(y)\right\}.$$

But this violates our previous observation that $|f(tx+(1-t)y)|\leq\max\left\{|f(x)|,|f(y)|\right\}$. This finishes the proof. ∎