7.12. Sequences of Differentiable FunctionsWe resume the subject from part 6.10 in which we found uniform convergence to be compatible with continuity. We won't get a similar result however with differentiability as we know, for example from the proof of the Weierstrass approximation theorem, that any function continuous on $[0,1]$ could be represented as the uniform limit of a sequence of polynomials, and thus as the uniform limit of a sequence of differentiable functions. (see [6.7.2]). The function $\mathrm{X}\frac{1}{2}$ e.g. is such a uniform limit on $[0,1]$ but is not differentiable at $\frac{1}{2}$. But even if the limit function proves to be differentiable there is no certainty that its derivative emanates from the derivatives of the sequence members. Due to the comparison test [6.10.7] we have for example
$\frac{1}{n}\mathrm{sin}\circ (n\mathrm{X})\underset{uf}{\to}0$,
as $\frac{1}{n}\mathrm{sin}(nx)\le \frac{1}{n}$ for all n and all x, but the sequence $((\frac{1}{n}\mathrm{sin}\circ n\mathrm{X}{)}^{\prime})=(\mathrm{cos}\circ n\mathrm{X})$ even fails to be pointwise convergent: $(\mathrm{cos}(n\pi ))=({(1)}^{n})$ for instance is divergent. Surprisingly the limit function will be differentiable with an appropriate derivative if the sequence of derivatives converges uniformly. The original sequence itself needs only to be pointwise convergent. Again, the mean value theorem is an essential argument in the proof, which is thus only valid for intervals, actually for bounded intervals in this case.
[7.12.2] is a local result which could be extended to a global version for $\mathcal{D}}^{1$functions and, in addition, for $\mathcal{C}}^{1$functions as well. Also, the restriction to bounded intervals is no longer needed.
