The right-hand inequality is clear, so we focus on the left-hand one. Given $v\in\mathrm{int}(\tau)$ and $e\in\mathcal{E}(\tau)$ we may write $v=\pi_{e}(v)+t_{0}\left(e-\pi_{e}(v)\right)$ for some $0<t_{0}<1$. Consider the restriction of $\varphi$ to the segment $[\pi_{e}(v),e]$, i.e. set $\theta(t):=\varphi\left(\pi_{e}(v)+t(e-\pi_{e}(v)\right)$, $t\in[0,1]$. If we denote by $\theta^{\prime}(t)$ the right-derivative of $\theta$ at $t$ then the convexity of $\theta$ yields

(A.2)

$$\theta^{\prime}(0)\leq\left(\theta(t_{0})-\theta(0)\right)/t_{0}$$

and

(A.3)

$$\theta^{\prime}(0)\leq\theta^{\prime}(t_{0})\leq\left(\theta(1)-\theta(t_{0})\right)/(1-t_{0}).$$

Now, by definition, $\theta^{\prime}(0)=D_{\pi_{e}(v)}\varphi(e)$, $t_{0}\theta^{\prime}(0)=D_{\pi_{e}(v)}\varphi(v)$ and $(1-t_{0})\theta^{\prime}(t_{0})=D_{v}\varphi(e)$, so that (A.2) reads

$$t_{0}D_{\pi_{e}(v)}\varphi(e)\leq\varphi(v)-\varphi(\pi_{e}(v)).$$

Since we also have $\sup_{\tau}\varphi=\sup_{\partial\tau}\varphi$ by convexity, this shows that

$$\|\varphi\|_{C^{0}(\tau)}\leq\|\varphi\|_{C^{0}(\partial\tau)}+\sup_{e\in\mathcal{E}(\tau),v\in\mathrm{int}(\tau)}\left|D_{\pi_{e}(v)}\varphi(e)\right|.$$

On the other hand, (A.2) combined with (A.3) yields

$$(1-t_{0})D_{\pi_{e}(v)}\varphi(e)\leq D_{v}\varphi(e)\leq\varphi(e)-\varphi\left(\pi_{e}(v)\right)-t_{0}D_{\pi_{e}(v)}\varphi(e)$$

and we conclude by Lemma A.2 below. ∎