have been chosen in order to remain as close as possible to the notations of refs [3–5]. We now calculate the expression of δT ( t ) that one has to introduce in equation (14) in order to obtain the nonlinear response in equation (15). We just consider the simple heating effect, when the dynamical heterogeneities are not considered: The supercooled liquid is thus just characterized by its thermal conductivity κ$_{th}$ and its total specific heat c . As in ref. [6], we consider that c is frequency dependent due to the fact that the slow degrees of freedom cannot contribute to c for frequencies much larger than f$_{α}$ = 1 /τ . Let us define ( x, y ) as the plane of our copper electrodes, with z = 0 for the lower electrode and z = L for the upper one. Due to their very high thermal conductivity and to their large thickness (6 mm), the two electrodes can be considered, to a very good approximation, as a thermostat. The temperature increase δθ ( x, y, z, t ) of the supercooled liquid at point ( x, y, z ) and time t thus vanishes for z = 0 and z = L . As the diameter D = 2 cm of the electrodes is typically one thousand times larger than L , we may consider that δθ does not depend on ( x, y ). We obtain δθ ( z, t ) by solving the heat propagation equation: By averaging spatially the solution of equation (16), we obtain the δT ( t ) to be used in equation (14): where τ$_{th}$ = cL $^{2}$/ ( κ$_{th}$π $^{2}$) and φ$_{2}$ = φ + arctan(2 ωτ$_{th}$ ). By using these expressions in equa- tion (14) and identifying it with equation (15), we obtain the upper limit to the heating contribution to χ$_{3}$ (dotted line in the Fig. 4 of the Letter). We now move to the problem evoked above, namely the fact that the finite relaxation time τ of the dipoles which contribute to the dielectric susceptibility should average out the modification of this susceptibility due to the oscillation δT$_{2}$ ( t ), specially in the case ωτ ≥ 1. As a consequence, the heating contribution to χ$_{3}$ should be reduced by a factor R ( ωτ ) ≤ 1. For a precise calculation of R ( ωτ ), one should replace Eq. (13) by a microscopic equation accounting for the dynamics of the dipoles in the case of a thermal bath where the 19 δT$_{0}$ = ϵ$_{0}$χ ′′ $_{1}$ωE$^{2}$L 2 24 κ$_{th}$ δT$_{2}$ ( t ) = δT$_{0}$ cos(2 ωt − φ$_{2}$ ) √ 1 + (2 ωτ$_{th}$ ) $_{2}$, (17) c ∂δθ ( z, t ) ∂t = κ$_{th}$ ∂ $^{2}$δθ ( z, t ) ∂z 2 + p ( t ) . (16)