Thanks for noting this. If in some case there is a positive level of capabilities for which P is 1, then we can just say that the level of capabilities denoted by C = 0 is the maximum level at which P is still 1. What will sort of change is that the constraint will be not C ≥ 0 but C ≥ (something negative), but that doesn’t really matter since here you’ll never want to set C<0 anyway. I’ve added a note to clarify this.
Maybe a thought here is that, since there is some stretch of capabilities along which P=1, we should think that P(.) is horizontal around C=0 (the point at which P can start falling from 1) for any given S, and that this might produce very different results from the e−CS example in which there would be a kink at C=0. But no—the key point is whether increases to S change the curve in a way that widens as C moves to the right, and so “act as price decreases to C”, not the slope of the curve around C=0. E.g. if P=1−C2/S (for C∈[0,√S], and 0 above), then in the k=0 case where the lab is trying to maximize (1−C2/S)C, they set C=√S/3, and so P is again fixed (here, at 2⁄3) regardless of S.
Thanks for noting this. If in some case there is a positive level of capabilities for which P is 1, then we can just say that the level of capabilities denoted by C = 0 is the maximum level at which P is still 1. What will sort of change is that the constraint will be not C ≥ 0 but C ≥ (something negative), but that doesn’t really matter since here you’ll never want to set C<0 anyway. I’ve added a note to clarify this.
Maybe a thought here is that, since there is some stretch of capabilities along which P=1, we should think that P(.) is horizontal around C=0 (the point at which P can start falling from 1) for any given S, and that this might produce very different results from the e−CS example in which there would be a kink at C=0. But no—the key point is whether increases to S change the curve in a way that widens as C moves to the right, and so “act as price decreases to C”, not the slope of the curve around C=0. E.g. if P=1−C2/S (for C∈[0,√S], and 0 above), then in the k=0 case where the lab is trying to maximize (1−C2/S)C, they set C=√S/3, and so P is again fixed (here, at 2⁄3) regardless of S.