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Proposition 17.4.969A8D9FFB919

Proposition 17.4.9.69A8D9FFB919 For $u,v\in \mathcal E^1(X,\theta ,\phi )$ and $u_ t=(1-t)u+tv$, the function $t\mapsto E_\theta ^\phi (u_ t)$ is continuously differentiable on $[0,1]$, with one-sided derivatives at the endpoints, and

\[ \frac{\mathrm d}{\mathrm dt}E_\theta ^\phi (u_ t) =\int _ X(v-u)\, \mu _{u_ t}. \]

The derivative is a polynomial in $t$.

Proof. By lemma 17.3.18 every integral $\int _ X(v-u)\langle \theta _ v^ a\wedge \theta _ u^{n-a}\rangle $ is finite. Expanding $\mu _{u_ t}$ by multilinearity shows that the right side is a polynomial in $t$ with these integrals as coefficients. The potentials $u_ t$ lie in $\mathcal E^1(X,\theta ,\phi )$ by corollary 17.3.17. Apply the cocycle formula of proposition 17.3.19 to $u_{t+s}$ and $u_ t$, with $s\neq 0$ and $t,t+s\in [0,1]$, and use $u_{t+s}-u_ t=s(v-u)$:

\[ \frac{E_\theta ^\phi (u_{t+s})-E_\theta ^\phi (u_ t)}s =\frac1{n+1}\sum _{j=0}^ n\int _ X(v-u)\langle \theta _{u_{t+s}}^ j\wedge \theta _{u_ t}^{n-j}\rangle . \]

Each mixed integral on the right is a polynomial in $s$ with the finite coefficients above, and at $s=0$ each of the $n+1$ terms equals $\int _ X(v-u)\mu _{u_ t}$. Letting $s\to 0$ proves the formula for the derivative, including the one-sided derivatives at the endpoints, and its continuity. $\square$

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