MATH.21:5.3 - Preserve values and integrals, then inspect differentiation

For n≥1, let f_n(x)=sin(nx)/n on [-1,1]. The inequality |f_n(x)|≤1/n holds for every input in the interval. Thus the functions converge uniformly to f(x)=0.

For a requested value error epsilon, n≥1/epsilon is sufficient. Integration over the interval is also controlled: the absolute integral of f_n-f is at most 2/n, by the interval length and the uniform bound. This estimate supplies an integral-error result without depending on cancellation of positive and negative values.

Now ask for the derivative at zero. Each f_n has derivative cos(nx), so f_n’(0)=1 for every n. The limit function has f’(0)=0. The value convergence therefore fails to justify passing this operation through the limit, even at that one point.

One possible sufficient repair, for differentiable functions on a common open interval, is pointwise convergence of the functions together with locally uniform convergence of their derivatives. A suitable limit theorem then identifies the limiting derivative. The present f_n fail that condition. If the construction can change, a family with controlled derivatives can support the stronger use; if it must stay fixed, obtain the derivative of its limit by another argument.

The result retains uniform convergence and its finite value and integral bounds. Only the derivative interchange has been refused. This is why the requested next operation belongs in the initial choice of approximation.