Section 8.2 Interchanging Limits

Limits do not commute automatically. Even when every object in sight is continuous or integrable, taking one limit can destroy the hypotheses needed for another. Uniform convergence is the condition that makes many familiar limit operations legitimate.

Subsection 8.2.1

Example 8.2.1.

Define

\begin{equation*} a_{n,k}=\frac{n}{n+k}. \end{equation*}

Then for each fixed \(n\text{,}\) \(\lim_{k\to\infty} a_{n,k}=0\text{,}\) so

\begin{equation*} \lim_{n\to\infty}\left(\lim_{k\to\infty} a_{n,k}\right)=0. \end{equation*}

But for each fixed \(k\text{,}\) \(\lim_{n\to\infty} a_{n,k}=1\text{,}\) and therefore

\begin{equation*} \lim_{k\to\infty}\left(\lim_{n\to\infty} a_{n,k}\right)=1. \end{equation*}

So the two iterated limits are different.

Theorem 8.2.2. Continuity of the Uniform Limit.

Let \(S \subseteq \R\text{,}\) let \(x_0 \in S\text{,}\) and suppose \(f_n \colon S \to \R\) is continuous at \(x_0\) for every \(n\text{.}\) If \(f_n \to f\) uniformly on \(S\text{,}\) then \(f\) is continuous at \(x_0\text{.}\)

Proof.

Let \(\varepsilon \gt 0\text{.}\) By uniform convergence, there exists \(N\) such that

\begin{equation*} |f_N(x)-f(x)| \lt \frac{\varepsilon}{3} \qquad \text{for all } x \in S. \end{equation*}

Since \(f_N\) is continuous at \(x_0\text{,}\) there exists \(\delta \gt 0\) such that

\begin{equation*} |f_N(x)-f_N(x_0)| \lt \frac{\varepsilon}{3} \qquad \text{whenever } x \in S \text{ and } |x-x_0| \lt \delta. \end{equation*}

For such \(x\text{,}\)

\begin{equation*} |f(x)-f(x_0)| \le |f(x)-f_N(x)| + |f_N(x)-f_N(x_0)| + |f_N(x_0)-f(x_0)| \lt \varepsilon. \end{equation*}

Therefore \(f\) is continuous at \(x_0\text{.}\)

Corollary 8.2.3.

The uniform limit of continuous functions is continuous.

Corollary 8.2.4.

Let \(S \subseteq \R\text{.}\) If each \(f_n \colon S \to \R\) is uniformly continuous on \(S\) and \(f_n \to f\) uniformly on \(S\text{,}\) then \(f\) is uniformly continuous on \(S\text{.}\)

Proof.

Let \(\varepsilon \gt 0\text{.}\) Choose \(N\) such that \(|f(x)-f_N(x)| \lt \varepsilon/3\) for all \(x \in S\text{.}\) Since \(f_N\) is uniformly continuous, there exists \(\delta \gt 0\) such that \(|f_N(x)-f_N(y)| \lt \varepsilon/3\) whenever \(x,y \in S\) and \(|x-y| \lt \delta\text{.}\) For such \(x,y\text{,}\)

\begin{equation*} |f(x)-f(y)| \le |f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)| \lt \varepsilon. \end{equation*}

Hence \(f\) is uniformly continuous on \(S\text{.}\)

Example 8.2.5.

The sequence \(f_n(x)=x^n\) on \([0,1]\) converges pointwise to the discontinuous limit described in Example 8.1.1. Thus continuity is not preserved by pointwise convergence.

Theorem 8.2.6. Integration of the Uniform Limit.

Let \(f_n \in R[a,b]\) for every \(n\text{,}\) and suppose \(f_n \to f\) uniformly on \([a,b]\text{.}\) Then \(f \in R[a,b]\text{,}\) and

\begin{equation*} \int_a^b f = \lim_{n\to\infty} \int_a^b f_n. \end{equation*}

Proof.

Let

\begin{equation*} c_n=\sup_{x \in [a,b]} |f(x)-f_n(x)|. \end{equation*}

Then \(c_n \to 0\text{.}\) Fix \(\varepsilon \gt 0\text{,}\) and choose \(n\) so that \(c_n \lt \varepsilon/(3(b-a))\text{.}\) Since \(f_n\) is integrable, the Darboux criterion (Proposition 7.1.7) gives a partition \(P\) of \([a,b]\) such that

\begin{equation*} U(P,f_n)-L(P,f_n) \lt \frac{\varepsilon}{3}. \end{equation*}

On each subinterval \(I_j\) of \(P\text{,}\) let \(M_j^n,m_j^n\) be the supremum and infimum of \(f_n\text{,}\) and let \(M_j,m_j\) be the corresponding quantities for \(f\text{.}\) From \(|f-f_n| \le c_n\) we obtain

\begin{equation*} M_j \le M_j^n+c_n \qquad\text{and}\qquad m_j \ge m_j^n-c_n. \end{equation*}

Summing over the partition gives

\begin{equation*} U(P,f)-L(P,f) \le U(P,f_n)-L(P,f_n)+2c_n(b-a) \lt \varepsilon. \end{equation*}

Therefore \(f\) is integrable.

Since \(f-f_n\) is now integrable, part (4) of Theorem 7.2.1 yields

\begin{equation*} \left|\int_a^b f-\int_a^b f_n\right| = \left|\int_a^b (f-f_n)\right| \le c_n(b-a). \end{equation*}

The right-hand side tends to \(0\text{,}\) so \(\int_a^b f_n \to \int_a^b f\text{.}\)

Example 8.2.7.

For each \(n \in \N\text{,}\) define

\begin{equation*} f_n(x)= \begin{cases} 4n^2x & \text{if } 0 \le x \le \frac{1}{2n},\\ 4n-4n^2x & \text{if } \frac{1}{2n} \lt x \le \frac{1}{n},\\ 0 & \text{if } \frac{1}{n} \lt x \le 1. \end{cases} \end{equation*}

Each \(f_n\) is continuous on \([0,1]\) and therefore integrable. Geometrically, the graph is a triangle of base \(1/n\) and height \(2n\text{,}\) so

\begin{equation*} \int_0^1 f_n = \frac12 \cdot \frac{1}{n} \cdot 2n = 1. \end{equation*}

For every fixed \(x \gt 0\text{,}\) we have \(f_n(x)=0\) for all sufficiently large \(n\text{,}\) and clearly \(f_n(0)=0\text{.}\) Thus \(f_n \to 0\) pointwise on \([0,1]\text{,}\) but

\begin{equation*} \int_0^1 \lim_{n\to\infty} f_n = 0 \neq 1 = \lim_{n\to\infty} \int_0^1 f_n. \end{equation*}

So pointwise convergence is not enough to interchange limit and integral.

Theorem 8.2.8. Differentiation Under the Limit Sign.

Let \(I \subseteq \R\) be an open interval, and let \(f_n \in C^1(I)\) for every \(n\text{.}\) Suppose there exists a point \(a \in I\) such that the numerical sequence \((f_n(a))\) converges, and suppose that \(f_n'\) converges uniformly on every compact interval \(K \subseteq I\) to a function \(g\text{.}\) Then there exists a differentiable function \(f \colon I \to \R\) such that \(f_n \to f\) uniformly on every compact interval of \(I\text{,}\) and

\begin{equation*} f'(x)=g(x) \qquad \text{for every } x \in I. \end{equation*}

Proof.

Let \(L=\lim_{n\to\infty} f_n(a)\text{.}\) Fix a compact interval \(K=[u,v] \subseteq I\text{,}\) and set \(\alpha=\min\{a,u\}\) and \(\beta=\max\{a,v\}\text{.}\) Then \([\alpha,\beta] \subseteq I\text{,}\) and by hypothesis \(f_n' \to g\) uniformly on \([\alpha,\beta]\text{.}\) Since each \(f_n'\) is continuous, Corollary 8.2.3 shows that \(g\) is continuous on \([\alpha,\beta]\text{.}\)

For each \(x \in K\text{,}\) the Fundamental Theorem of Calculus gives

\begin{equation*} f_n(x)=f_n(a)+\int_a^x f_n'(t)\,dt. \end{equation*}

Define

\begin{equation*} f(x)=L+\int_a^x g(t)\,dt \qquad (x \in I). \end{equation*}

Then for \(x \in K\text{,}\)

\begin{equation*} \begin{aligned} |f_n(x)-f(x)| &\le |f_n(a)-L| + \left|\int_a^x \bigl(f_n'(t)-g(t)\bigr)\,dt\right| \\ &\le |f_n(a)-L| \\ &\qquad + (\beta-\alpha) \sup_{t \in [\alpha,\beta]} |f_n'(t)-g(t)|. \end{aligned} \end{equation*}

The right-hand side tends to \(0\text{,}\) uniformly in \(x \in K\text{.}\) Hence \(f_n \to f\) uniformly on \(K\text{.}\)

Because \(g\) is continuous on every compact interval of \(I\text{,}\) the Fundamental Theorem of Calculus implies that \(f\) is differentiable on \(I\) and that \(f'(x)=g(x)\) for every \(x \in I\text{.}\)

Example 8.2.9.

Uniform convergence of \(f_n\) alone does not justify differentiating term by term. Let

\begin{equation*} f_n(x)=\frac{\sin(nx)}{n} \qquad (x \in \R). \end{equation*}

Since \(|f_n(x)| \le 1/n\text{,}\) we have \(f_n \to 0\) uniformly on \(\R\text{.}\) But

\begin{equation*} f_n'(x)=\cos(nx), \end{equation*}

and the sequence \((\cos(nx))\) does not even converge pointwise on \(\R\text{;}\) for example, \(\cos(n\pi)=(-1)^n\) oscillates.

Theorems Theorem 8.2.6 and Theorem 8.2.8 explain why power series can be integrated and differentiated term by term on closed subintervals inside their interval of convergence.

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