2.7 Uniform Convergence
When two limiting operations are performed on functions, the order in which they are carried out can make a difference. For example, let \[ f_n(x) = \begin {cases} 0, & x < 0,\\ nx, & 0\leq x \leq \frac {1}{n},\\ 1, & x> \frac {1}{n}. \end {cases} \] For each fixed \(x\), \(f_n(x)\longrightarrow f(x)\) as \(n\longrightarrow \infty \), where \[f(x) = \begin {cases} 0, & x\leq 0,\\ 1, & x>0. \end {cases}\]
Now \begin {gather*} \lim _{n\rightarrow \infty }\, \lim _{x\rightarrow 0^+} f_n(x) = \lim _{n \rightarrow \infty } 0 = 0,\\ \text {but}\qquad \lim _{x\rightarrow 0^+}\,\lim _{n\rightarrow \infty } f_n(x) = \lim _{x \rightarrow 0^+} f(x) = 1 . \end {gather*} Note also that every \(f_n\) is continuous everywhere and \(f_n(x) \longrightarrow f(x)\) for each \(x\), but the limit function \(f\) is discontinuous at \(x= 0\).
Example 2.7.1. Let \[ f_n(x) = \begin {cases} n e^{-nx}, & x>0,\\ 0, & x\leq 0. \end {cases} \] Then \(f_n(x) \longrightarrow 0\) for every \(x\), so \(f_n\) converges pointwise to \(f(x) = 0\). Now \[\int ^{\infty }_0f_n(x)\,dx = \int ^{\infty }_0 n e^{-nx}\,dx = 1 \quad \text {for every } n, \qquad \text {but}\qquad \int ^{\infty }_0 f(x)\, dx = 0 .\] Thus, although \(f_n \longrightarrow f\) pointwise, \(\int ^{\infty }_0f_n(x)\, dx\) does not converge to \(\int ^{\infty }_0f(x)\, dx\). In other words, the limit of the integrals of \(f_n\) is not the integral of the limit of \(f_n\).
Example 2.7.2. Let \(f_n(x) = x^n\) on \([0,1]\). Find the pointwise limit, and show that the convergence is uniform on \([0,a]\) for each \(a<1\) but not on \([0,1]\).
Solution. For \(0\leq x<1\), \(x^n\longrightarrow 0\), and \(f_n(1) = 1\). So the limit is \(f(x) = 0\) for \(x<1\) and \(f(1) = 1\), which is discontinuous at \(1\). Since each \(f_n\) is continuous, the convergence cannot be uniform on \([0,1]\) (by the theorem on continuity of uniform limits in the next section). Directly, \(\sup _{0\leq x<1}|x^n - 0| = 1\) for every \(n\).
On \([0,a]\) with \(a<1\), \(\sup _{0\leq x\leq a}|x^n| = a^n\longrightarrow 0\), so the convergence is uniform. For example, with \(a = 0.9\) and \(\varepsilon = 0.01\), any \(n\geq 44\) works, since \(0.9^{44}\approx 0.0097\).
Example 2.7.3. Let \(f_n(x) = \frac {\sin nx}{\sqrt {n}}\) on \(\mathbb {R}\). Show that \(f_n\longrightarrow 0\) uniformly, but that the derivatives \(f_n'\) do not converge to the derivative of the limit.
Solution. Since \(|f_n(x)|\leq \frac {1}{\sqrt {n}}\) for all \(x\), \(\sup _x|f_n(x)|\longrightarrow 0\) and the convergence is uniform. The limit \(f = 0\) has \(f' = 0\). But \(f_n'(x) = \sqrt {n}\cos nx\), and at \(x = 0\) this equals \(\sqrt {n}\longrightarrow \infty \). So uniform convergence of \(f_n\) does not allow us to differentiate term by term; extra conditions on \(f_n'\) are needed.
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