Log and Brier losses are strictly proper

If $t>1$, then $1\mathrm{/}u<1$ for $1<u\le t$, so [F2] gives $\log t<t-1$. If $0<t<1$, then $1\mathrm{/}u>1$ for $t\le u<1$, so $\log t=-{\int }_t^{1}du\mathrm{/}u<-(1-t)=t-1$. At $t=1$ equality holds. Therefore $-\log t\ge 1-t$ for every $t>0$, with equality exactly at $t=1$.

If $p_j>0$ and $q_j=0$ for some $j$, then [F1] gives ${\bar{S}}_{\log }(p,q)=+\mathrm{\infty }$, while ${\bar{S}}_{\log }(p,p)=-{\sum }_{j:p_j>0}{p}_j\log p_j<\mathrm{\infty }$ because this is a finite sum of finite terms. Thus propriety is strict in this case.

Otherwise $q_j>0$ for every $j$ with $p_j>0$. The truthful risk is finite, and cancellation on the support of $p$ gives ${\bar{S}}_{\log }(p,q)-{\bar{S}}_{\log }(p,p)={\sum }_{j:p_j>0}{p}_j\log \frac{p_j}{q_j}\mathrm{.}$ Applying step 1.1 to $t=q_j\mathrm{/}{p}_j$ yields ${\sum }_{j:p_j>0}{p}_j\log \frac{p_j}{q_j}\ge 1-{\sum }_{j:p_j>0}{q}_j\ge 0.$ If the displayed difference is zero, both nonnegative gaps in this chain are zero. Equality in step 1.1 then forces $q_j\mathrm{/}{p}_j=1$ on the support of $p$, and the second gap forces $q$ to have no mass outside that support. Hence $q=p$. This proves strict propriety for the logarithmic loss.

Expanding the Brier loss and using ${\sum }_j{p}_j=1$ gives ${\bar{S}}_{\mathrm{B}}(p,q)={\sum }_i{q}_i^2-2{\sum }_i{p}_i{q}_i+1,$ whereas ${\bar{S}}_{\mathrm{B}}(p,p)=1-{\sum }_i{p}_i^2$. Subtraction yields ${\bar{S}}_{\mathrm{B}}(p,q)-{\bar{S}}_{\mathrm{B}}(p,p)={\sum }_i(q_i-p_i{)}^2\ge 0$, with equality exactly when every $q_i=p_i$. Thus the Brier loss is strictly proper as well.