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Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-2u_s} \Delta_{\mathbb{S}^2} u_s-2. $$ Linearizing at $s=0$, we see that the linearized $\phi=\frac{d}{ds}|_{s=0} \phi_s$ from your Fact 2 must satisfy $$ \phi=-2 \Delta_{\mathbb{S}^2} v-4 $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor. This means $\phi$ is not arbitary but is in the image of the operator $\Delta_{\mathbb{S}^2} +2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.

Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-2u_s} \Delta_{\mathbb{S}^2} u_s-2\frac{4\pi}{\int_{\mathbb{S}^2}e^{-2u_s} dvol_{\mathbb{S}_2}}. $$ Linearizing at $s=0$, we see that the linearized $\phi=\frac{d}{ds}|_{s=0} \phi_s$ from your Fact 2 must satisfy $$ \phi=-2 \Delta_{\mathbb{S}^2} v-4v $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor (we used that the average scalar curvature is fixed so $v$ has average zero in order to eliminate the last term). This means $\phi$ is not arbitrary, but is in the image of the operator $\Delta_{\mathbb{S}^2} +2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.

Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-u_s} \Delta_{\mathbb{S}^2} u_s. $$ Linearizing at $s=0$, we see that the linearized $\phi$ from your Fact 2 must satisfy $$ \phi=2-2 \Delta_{\mathbb{S}^2} v $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor. This means $\phi$ is not arbitary but is in the image of the operator $\Delta_{\mathbb{S}^2} -2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.

Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-2u_s} \Delta_{\mathbb{S}^2} u_s-2. $$ Linearizing at $s=0$, we see that the linearized $\phi=\frac{d}{ds}|_{s=0} \phi_s$ from your Fact 2 must satisfy $$ \phi=-2 \Delta_{\mathbb{S}^2} v-4 $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor. This means $\phi$ is not arbitary but is in the image of the operator $\Delta_{\mathbb{S}^2} +2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.

Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely varying, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the metric. In otherwords, what you can freely vary is the metric and if you do so (within a fixed conformal class) you see that

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-u_s} \Delta_{\mathbb{S}^2} u_s. $$ Linearizing at $s=0$, we see that the linearized $\phi$ from your Fact 2 must satisfy $$ \phi=2-2 \Delta_{\mathbb{S}^2} v $$ where $v=\frac{d}{ds}|_{s=0}$ is the variation of the conformal factor. This means $\phi$ is not arbitary but is in the image of the operator $\Delta_{\mathbb{S}^2} -2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-u_s} \Delta_{\mathbb{S}^2} u_s. $$ Linearizing at $s=0$, we see that the linearized $\phi$ from your Fact 2 must satisfy $$ \phi=2-2 \Delta_{\mathbb{S}^2} v $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor. This means $\phi$ is not arbitary but is in the image of the operator $\Delta_{\mathbb{S}^2} -2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.