(Heuristic) Consider the Harder-Narasimhan decomposition (4) of $L_{1}$:

$$0=\mathcal{E}_{0}\to\mathcal{E}_{1}\to\ldots\to\mathcal{E}_{N}=L_{1},$$

fitting into the distinguished triangles

$$\mathcal{E}_{i-1}\to\mathcal{E}_{i}\to L_{i}^{\prime}\to\mathcal{E}_{i-1}[1],$$

where $L_{i}^{\prime}$ represents an object in $\mathcal{P}(\phi_{i})$, with $\phi_{1}>\phi_{2}\ldots>\phi_{N}$. Since by assumption $L_{1}$ is almost calibrated, we have $L_{1}\in\mathcal{P}(-\frac{1}{2}<\phi<\frac{1}{2})$, hence $-\frac{1}{2}<\phi_{i}<\frac{1}{2}$. Since the central charges satisfy

$$Z(L_{1})=\sum_{1}^{N}Z(L_{i}^{\prime}),\quad\arg Z(L_{i})=\pi\phi_{i},$$

we must have $\hat{\theta}_{1}\leq\pi\phi_{1}$. The conjectural description of the Bridgeland stability condition requires that $L_{i}^{\prime}$ has a special Lagrangian representative with constant Lagrangian phase $\pi\phi_{i}$. A weaker requirement which suffices for us is that there exists a representative $L_{i}^{\prime}$ with Lagrangian angle function $\theta_{L_{i}^{\prime}}$ satisfying the oscillation bound $|\theta_{L_{i}^{\prime}}-\pi\phi_{i}|<\epsilon$ for any given $\epsilon>0$. It is expected that this flexibility allows one to assume sufficient smoothness on the Lagrangian.

By combining the distinguished triangles, we obtain a new distinguished triangle

$$L_{1}^{\prime}\to L\to L^{\prime\prime}\to L_{1}^{\prime}[1].$$

Here $L_{1}^{\prime},L,L^{\prime\prime}$ are all almost calibrated. Suppose for contradiction that $\sup_{L}\theta_{L}<\hat{\theta}_{1}$. Then $\sup_{L}\theta_{L}<\pi\phi_{1}$, and we can arrange $\sup_{L}\theta_{L}<\inf_{L_{1}^{\prime}}\theta_{L_{1}^{\prime}}$. The Floer degree formula (63) implies $CF^{0}(L_{1}^{\prime},L)=0$, and in particular $HF^{0}(L_{1}^{\prime},L)=0$. The distinguished triangle splits: $L^{\prime\prime}\simeq L\oplus L_{1}^{\prime}[1]$. Since $L^{\prime\prime}$ is almost calibrated, it lies in $\mathcal{P}(-\frac{1}{2}<\phi<\frac{1}{2})$, and so must $L_{1}^{\prime}[1]$. But $L_{1}^{\prime}\in\mathcal{P}(\phi_{1})$ implies $L_{1}^{\prime}[1]\in\mathcal{P}(\phi_{1}+1)$. Since $\phi_{1}+1>\frac{1}{2}$, we know $\mathcal{P}(\phi_{1}+1)\cap\mathcal{P}(-\frac{1}{2}<\phi<\frac{1}{2})=\emptyset$, contradiction. This proves $\sup_{L}\theta_{L}\geq\hat{\theta}_{1}$, subject to the conjectural existence of the Bridgeland stability condition.

A very similar argument, beginning with the Harder-Narasimhan decomposition of $L_{2}$, would show $\inf_{L}\theta_{L}\leq\hat{\theta}_{2}$. ∎