Function Values and Slopes
Uniform convergence controls the values of functions everywhere on their domain. Differentiation, however, measures how function values change over arbitrarily short distances. A small error in the values does not necessarily mean a small error in the slopes: an error can be tiny in height and still change rapidly over a narrow region.
This difference is important when studying a sequence of differentiable functions \(f_n\) and its uniform limit \(f\). Uniform convergence alone does not guarantee that \(f\) is differentiable, or that the derivatives \(f_n'\) converge. We will see both failures explicitly. We will also prove a useful positive result: a common bound on the derivatives passes to the limit as a Lipschitz bound. The Uniform Derivative Convergence Criterion from earlier in this course gives a stronger sufficient condition for differentiability of the limit; here we focus on why additional control is needed.
A derivative bound limits how steep any member of the sequence can be. The next theorem explains what that restriction implies for a uniform limit. Notice that its conclusion is a Lipschitz estimate, not differentiability: a Lipschitz function can still have a corner.
A Common Derivative Bound Controls the Limit
Proof. Choose any \(x,y\in I\) with \(x<y\). Each \(f_n\) is continuous on \([x,y]\) and differentiable on \((x,y)\). The Mean Value Theorem therefore gives a point \(c_n\in(x,y)\) such that $$ f_n(y)-f_n(x)=f_n'(c_n)(y-x). $$ Taking absolute values and using the derivative bound yields $$ |f_n(y)-f_n(x)|\leq M(y-x). $$ Uniform convergence implies pointwise convergence at both \(x\) and \(y\), so \(f_n(y)-f_n(x)\to f(y)-f(x)\). Absolute value is continuous, and hence taking limits in the inequality gives $$ |f(y)-f(x)|\leq M(y-x). $$ If \(x=y\), the desired inequality reads \(0\leq0\). If \(y<x\), apply the result just proved to the ordered pair \(y<x\), obtaining \(|f(x)-f(y)|\leq M(x-y)\). Thus the estimate holds for all \(x,y\in I\), proving that \(f\) is Lipschitz with constant \(M\). \(\square\)
The proof uses the Mean Value Theorem before passing to the limit. It does not attempt to pass a limit through \(f_n'\); no convergence of the derivatives was assumed. A uniform derivative bound controls the change between any two inputs, and uniform convergence preserves that two-point inequality.
Worked Example: A Uniform Limit with a Corner
For \(n\geq1\), define $$ f_n(x)=\sqrt{x^2+\frac1n} \qquad (-1\leq x\leq1). $$ Each \(f_n\) is differentiable, with $$ f_n'(x)=\frac{x}{\sqrt{x^2+1/n}}. $$ We first check uniform convergence to \(f(x)=|x|\). For every \(x\), $$ 0\leq f_n(x)-|x| =\frac{1/n}{\sqrt{x^2+1/n}+|x|} \leq\frac1{\sqrt n}. $$ The last bound also follows directly from \(\sqrt{x^2+1/n}\leq |x|+1/\sqrt n\); alternatively, the difference is largest at \(x=0\), where it equals \(1/\sqrt n\). Since \(1/\sqrt n\to0\), the convergence is uniform.
The limit is not differentiable at \(0\). Indeed, its difference quotient there is $$ \frac{f(h)-f(0)}{h}=\frac{|h|}{h}, $$ which equals \(1\) for \(h>0\) and \(-1\) for \(h<0\). The one-sided limits disagree. Thus a uniform limit of differentiable functions need not be differentiable. In this example, the derivatives also satisfy \(|f_n'(x)|\leq1\), so the Derivative-Bound Stability Theorem applies and correctly shows that the limit is \(1\)-Lipschitz. Being Lipschitz does not rule out the corner.
For each fixed \(x>0\), \(f_n'(x)\to1\), and for each fixed \(x<0\), \(f_n'(x)\to-1\). At \(x=0\), \(f_n'(0)=0\) for every \(n\). These pointwise limits do not make \(f\) differentiable at zero. In particular, the limiting behavior of slopes on either side cannot supply a derivative at a point where the difference quotient itself has no limit.
Small Uniform Errors Can Have Large Slopes
The preceding example shows that differentiability may fail even when the approximating derivatives are bounded. There is a different failure as well: the functions can converge uniformly while their derivatives do not even converge pointwise. The height of a function and the rate at which that height changes are separate quantities.
Worked Example: Uniformly Small Oscillations
On \(\mathbb{R}\), let $$ g_n(x)=\frac{\sin(n^2x)}{n}. $$ Since \(|\sin(n^2x)|\leq1\), $$ |g_n(x)|\leq\frac1n \qquad (x\in\mathbb{R}). $$ Thus \(g_n\to0\) uniformly on the whole real line. But $$ g_n'(x)=n\cos(n^2x), \qquad\text{so}\qquad g_n'(0)=n. $$ At \(x=0\), the derivative values tend to infinity and therefore do not converge to a finite real number. The functions have uniformly vanishing height, while their slopes at the origin grow without bound.
This example shows why an estimate on \(\sup_x|f_n(x)-f(x)|\) cannot, by itself, estimate \(\sup_x|f_n'(x)-f'(x)|\). Differentiation magnifies rapid variation: here the oscillation frequency is \(n^2\), while the amplitude is only \(1/n\), and differentiating multiplies the oscillatory factor by \(n^2\).
Worked Example: Derivatives Converge Pointwise but Not Uniformly
Define \(u_n(x)=x+n x^2e^{-n^2x^2}\) on \(\mathbb{R}\). Set \(\phi(t)=t^2e^{-t^2}\), so the added term is \(\phi(nx)/n\). The function \(\phi\) is bounded: it is continuous, tends to zero as \(|t|\to\infty\), and has finite maximum. Consequently, $$ \sup_{x\in\mathbb{R}}|u_n(x)-x| =\frac1n\sup_{t\in\mathbb{R}}|\phi(t)| \longrightarrow0. $$ Thus \(u_n\to u\) uniformly, where \(u(x)=x\).
Differentiating the added term gives $$ u_n'(x)=1+\phi'(nx), \qquad \phi'(t)=2t(1-t^2)e^{-t^2}. $$ For each fixed \(x\neq0\), \(nx\) tends in absolute value to infinity, so \(\phi'(nx)\to0\); at \(x=0\), \(\phi'(0)=0\). Therefore \(u_n'(x)\to1=u'(x)\) at every fixed \(x\). The convergence is nevertheless not uniform. At \(x=1/n\), $$ |u_n'(1/n)-u'(1/n)| =|\phi'(1)| =0. $$ This point does not demonstrate failure, so choose instead \(x=1/(2n)\). Then $$ |u_n'(1/(2n))-u'(1/(2n))| =|\phi'(1/2)| =\frac{3}{4}e^{-1/4}>0. $$ The same positive discrepancy occurs for every \(n\), so the supremum of the derivative error cannot tend to zero. This example separates pointwise convergence of derivatives from uniform convergence: the discrepancy is concentrated in a region that narrows toward zero.
What to Check When Differentiating a Limit
These examples expose three different levels of control. Uniform convergence controls the function values. A common bound on derivatives additionally controls how quickly those values can change, which is why it forces a Lipschitz limit. But even that bound need not yield differentiability. Convergence of derivatives pointwise is also not the same as uniform convergence of derivatives, as the narrowing-region example demonstrates.
The Uniform Derivative Convergence Criterion established earlier in this course supplies a sufficient condition for identifying the derivative of a limit: the derivatives must converge uniformly, together with the appropriate convergence information for the functions. The examples here explain why a hypothesis of that kind is substantive rather than automatic. They do not say uniform convergence of derivatives is necessary in every situation; they show only that uniform convergence of the functions, by itself, is not enough.
Check Your Understanding
Use the estimates and examples in this tutorial to answer the following questions.
- Where in the proof of the Derivative-Bound Stability Theorem is the Mean Value Theorem used, and what estimate does it provide?
- Why is \(f(x)=|x|\) not differentiable at \(0\), even though it is a uniform limit of differentiable functions in the first worked example?
- For \(g_n(x)=\sin(n^2x)/n\), what are \(g_n(0)\) and \(g_n'(0)\)?
- In the narrowing-region example, why does a fixed \(x\neq0\) eventually lie outside the region of substantial derivative error?
- Does a common bound on the derivatives guarantee that the uniform limit is differentiable? Give the relevant example.