Convergence of Functions

Definition (Pointwise convergence)

A sequence of functions \(f_{n}:A \to \mathbb{R}\) converges pointwise to \(f:A \to \mathbb{R}\) if for every \(x \in A\) we have that \(\lim_{ n \to \infty } f_{n}(x)=f(x)\).

Example

Consider \(f_{n}(x) = x^{n}\) for \(x \in (-1,+\infty)\). Note that

\[\lim_{ n \to \infty } f_{n}(x) \begin{cases} 0 \quad \text{if } \lvert x \rvert <1 \\ 1 \quad \text{if } x=1 \\ \infty \quad \text{if } x>1 \end{cases}.\]

Let

\[f(x) = \begin{cases} 0 \quad \text{if } x \in (-1,1) \\ 1 \quad \text{if } x=1 \end{cases}.\]

Then \(f_{n}\) converges point by point to \(f\) on \((-1,1]\). How would we prove this from the definition?

Proof: Let \(\varepsilon>0\). Let \(N = \max \left\{ 1, \left\lfloor \frac{\ln(\varepsilon)}{\ln(\lvert x \rvert)} \right\rfloor -1 \right\}\). Then, for \(n\geq N\),

\[n > \frac{\ln(\varepsilon)}{\ln\lvert x \rvert } \iff n\ln \lvert x \rvert < \ln \varepsilon \iff \lvert x^{n} \rvert = \lvert x \rvert ^{n} < \varepsilon.\]

CAREFUL: Note that \(N = N(x, \varepsilon)\). If \(x\) changes, \(N\) changes. Moreover, \(\lim_{ x \to 1^{-} } N = +\infty\). That is, there is no \(N\) such that \(\lvert x^{n} \rvert < \varepsilon\) for every \(n>N\) and every \(x \in (-1, 1])\)

Definition (Uniform convergence)

A sequence of functions \(f_{n}:A\to \mathbb{R}\) converges uniformly to \(f:A\to \mathbb{R}\) if for every \(\varepsilon>0\) there exists \(N \in \mathbb{N}\) such that for \(n\geq N\) we have that, for every \(x \in A\),

\[\lvert f_{n}(x)-f(x) \rvert < \varepsilon.\]

Example

Consider now the same sequence of functions as in the previous example but with \(x \in (0,a)\), where \(a<1\). We will prove that it converges uniformly.

Proof: Let \(\varepsilon>0\). Let \(N = \max \left\{ 1, \left\lfloor \frac{\ln(\varepsilon)}{\ln(a)} \right\rfloor -1 \right\}\). Then, for \(n\geq N\) and for every \(x \in (0,a)\)

\[n > \frac{ln(\varepsilon)}{\ln (a)}\geq \frac{\ln(\varepsilon)}{\ln\lvert x \rvert } \iff n\ln \lvert x \rvert < \ln \varepsilon \iff \lvert x^{n} \rvert = \lvert x \rvert ^{n} < \varepsilon.\]

Definition (Supremum norm)

Let \(f:A\to \mathbb{R}\) be bounded. Define \(\lVert f \rVert_{\infty} := \sup \{ \lvert f(x) \rvert: x \in A \}\). Note that if \(\lVert f \rVert_{\infty} = 0\), then \(\lvert f(x) \rvert \leq 0\) for every \(x \in A\), and therefore \(f=0\).

On the other hand, if \(f:A\to \mathbb{R}\), \(g:A\to \mathbb{R}\) are bounded, then

\[\sup \{ \lvert f(x) + g(x) \rvert: x \in A \} \leq \sup \{ \lvert f(x) \rvert + \lvert g(x) \rvert : x \in A \} \leq \sup \{ \lvert f(x) \rvert: x \in A \} + \sup \{ \lvert g(x) \rvert: x \in A \}.\]

Hence \(\lVert f+g \rVert_{\infty} \leq \lVert f \rVert_{\infty} + \lVert g \rVert_{\infty}\). Therefore \(\lVert \rVert_{\infty}\) is a norm on the space of bounded functions \(A\to \mathbb{R}\).

Uniform convergence and the supremum norm

Let \(f_{n}:A\to \mathbb{R}\) converge uniformly to \(f:A\to \mathbb{R}\). Then, given \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that if \(n\geq N\), \(\lvert f_{n}(x) - f(x) \rvert < \varepsilon\) for every \(x \in A\). Hence

\[\sup \{ \lvert f_{n}(x) - f(x) \rvert : x \in A \} \leq \varepsilon,\]

so \(\lVert f_{n}-f \rVert_{\infty} \leq \varepsilon\) for \(n\geq N\).

Exercise

Let \(f_{n}:A\to \mathbb{R}\) and \(f:A\to \mathbb{R}\) be such that \(\lVert f_{n}-f \rVert_{\infty} \underset{n \rightarrow \infty}{\longrightarrow} 0\). Then \(f_{n}\) converges uniformly to \(f\).

Example

Consider \(f_{n}(x) = x^{n}(1-x)\) for \(x \in [0,1]\). Then \(f_{n} \underset{n \rightarrow \infty}{\longrightarrow} 0\) pointwise. We will prove that it converges uniformly. Let \(0 < \delta < \min \left\{ \varepsilon,1 \right\}\). Then, on \([0,1-\delta]\), we have that

\[\lvert x \rvert ^{n} \lvert 1-x \rvert < \varepsilon.\]

Now,

\[\begin{aligned} \lvert x^{n} \rvert \lvert 1-x \rvert &\leq \lvert x \rvert ^{n} < \varepsilon \\ \iff n > \left\lfloor \frac{\ln(\varepsilon)}{\ln(1-\delta)} \right\rfloor &= N. \end{aligned}\]

Hence, for \(n\geq N\), we have that \(\lvert x^{n}(1-x) \rvert < \varepsilon\) if \(x \in [0,1-\delta]\). On the other hand, if \(x \in [1-\delta,1]\), then \(\lvert x^{n}(1-x) \rvert \leq 1-x \leq \delta < \varepsilon\).

Theorem (Boundedness of the limit function)

Let \(f_{n}:A\to \mathbb{R}\) be bounded functions converging uniformly to \(f:A\to \mathbb{R}\). Then \(f\) is bounded.

Proof: Let \(\varepsilon = 1\). There exists \(N \in \mathbb{N}\) such that \(\lvert f_{n}(x) - f(x) \rvert < 1\) for every \(n\geq N\). Then

\[f_{n}(x) -1 < f(x) < f_{n}(x)+1\]

Since \(f_{n}\) is bounded (for any \(n\geq N\)), we know that there exists \(M \in \mathbb{R}\) such that \(-M \leq f_{n}(x) \leq M\). Hence \(-M-1 \leq f(x) \leq M+1\).

Example

Consider \(f_{n}(x) = \frac{1}{x+1/n}\); then \(\lvert f_{n}(x) \rvert \leq n\) for \(x \in [0,+\infty]\), but \(\lim_{ n \to \infty } f_{n}(x) = \frac{1}{x}\) is not bounded. Therefore the convergence is not uniform.

Example

Theorem (Uniform convergence of continuous functions)

Let \(f_{n}:A\to \mathbb{R}\) be continuous functions converging uniformly to \(f:A\to \mathbb{R}\). Then \(f\) is continuous.

Proof: Let \(\varepsilon>0\). There exists \(N \in \mathbb{N}\) such that if \(n\geq N\),

\[\lvert f_{n}(x) - f(x) \rvert < \frac{\varepsilon}{3}\]

for every \(x \in A\). Moreover, there exists \(\delta>0\) such that if \(\lvert x-x_{0} \rvert < \delta\), then \(\lvert f_{N}(x) - f_{N}(x_{0}) \rvert < \frac{\varepsilon}{3}\). Hence, if \(\lvert x-x_{0} \rvert < \delta\)

\[\begin{aligned} \lvert f(x)-f(x_{0}) \rvert &\leq \lvert f(x) - f_{N}(x) \rvert + \lvert f_{N}(x) - f(x_{0}) \rvert \\ &\leq \lvert f(x) - f_{N}(x) \rvert + \lvert f_{N}(x) - f_{N}(x_{0}) \rvert + \lvert f_{N}(x_{0}) - f_{N}(x) \rvert < \varepsilon. \end{aligned}\]

Conclude that \(f\) is continuous.

Example

Consider \(f_{n}(x) = x^{n}\), \(x \in [0,1]\). Note that

\[\lim_{ n \to \infty } f_{n}(x) = \begin{cases} 0 \quad \text{if } x \in [0,1) \\ 1 \quad \text{if } x = 1 \end{cases}.\]

Conclude that the convergence is not uniform.

Definition (Uniformly Cauchy)

Let \(f_{n}:A\to \mathbb{R}\) be a sequence of functions. We say that \(f_{n}\) is uniformly Cauchy if, given \(\varepsilon>0\), there exists \(N\) such that if \(n,m\geq N\) we have that \(\lvert f_{n}(x) - f_{m}(x) \rvert < \varepsilon\) for every \(x \in A\).

Lemma (Uniform convergence if and only if uniformly Cauchy)

Let \(f_{n}:A\to \mathbb{R}\) be a sequence of functions and \(f:A\to \mathbb{R}\). Then \(f_{n} \underset{n \rightarrow \infty}{\longrightarrow} f\) uniformly if and only if \(f_{n}\) is uniformly Cauchy.

Proof: (\(\implies\)): Let \(\varepsilon>0\). Then there exists \(N \in \mathbb{N}\) such that for \(n\geq N\) we have that \(\lvert f_{n}(x) - f(x) \rvert < \frac{\varepsilon}{2}\) for every \(x \in A\). Now for all \(n,m\geq N\),

\[\lvert f_{n}(x)-f_{m}(x) \rvert \leq \lvert f_{n}(x) - f(x)\rvert + \lvert f(x) - f_{m}(x) \rvert < \varepsilon.\]

Therefore it is uniformly Cauchy. (\(\impliedby\)): Let \(\varepsilon>0\). Then there exists \(N\) such that for all \(m,n\geq N\) we have that \(\lvert f_{n}(x)-f_{m}(x) \rvert < \frac{\varepsilon}{2}\)for every \(x \in A\). Let \(x_{0} \in A\). Then \(\{f_{n}(x_{0})\}_{n=1}^\infty\) is Cauchy and therefore convergent, i.e., there exists \(\ell \in \mathbb{R}\) such that \(f_{n}(x_{0}) \underset{n \rightarrow \infty}{\longrightarrow} \ell\). Define \(f(x) = \lim_{ n \to \infty } f_{n}(x)\). Then, given \(x \in A\), there exists \(N_{0}\geq N\) such that if \(n \geq N_{0}(x)\) then

\[\lvert f(x) - f_{n}(x) \rvert < \frac{\varepsilon}{2}.\]

Now, if \(m\geq N\),

\[\begin{aligned} \lvert f(x) - f_{m}(x) \rvert &\leq \lvert f(x) - f_{N_{0}}(x) \rvert + \lvert f_{N_{0}}(x) - f_{m}(x) \rvert \\ & \leq \frac{\varepsilon}{2} + \lvert f_{N_{0}}(x) - f_{m}(x) \rvert < \varepsilon \quad \text{since it is uniformly Cauchy.} \end{aligned}\]

Theorem (Uniform convergence and Riemann integrability)

Let \(f_{n}:[a,b] \to \mathbb{R}\) be Riemann integrable functions converging uniformly to \(f:[a,b] \to \mathbb{R}\). Then \(f\) is Riemann integrable and moreover

\[\lim_{ n \to \infty } \int_{a}^{b} f_{n}(x) \, dx = \int_{a}^{b} f(x) \, dx.\]

Proof: Given \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that for every \(n\geq N\), \(\lvert f_{n}(x) - f(x) \rvert < \frac{\varepsilon}{4(b-a)}\) for every \(x \in [a,b]\). Since \(f_{N}\) is Riemann integrable, there exists \(P_{\varepsilon}\) such that \(U(f_{N}, P_{\varepsilon}) - L(f_{N},P_{\varepsilon}) < \frac{\varepsilon}{2}\). Let \(P_{\varepsilon} = \{ a_{0}=a < \dots < a_{n} = b \}\). Then

\[(a_{i+1}-a_{i}) \sup \{ f(x): a_{i} \leq x \leq a_{i+1} \} \leq \left( \sup \{ f_{N}(x) : a_{i} \leq x \leq a_{i+1}\} + \frac{\varepsilon}{4(b-a)} \right)(a_{i+1}-a_{i}).\]

since \(f_{N}(x) - \varepsilon \leq f(x) \leq f_{N}(x) + \varepsilon\) for every \(x \in A\). In the same way,

\[\sup \{ f_{N}(x):a_{i}\leq x\leq a_{i+1} \} - \frac{\varepsilon}{4(b-a)} \leq \sup \{ f(x):a_{i}\leq x\leq a_{i+1} \}.\]

Then

\[U(f, P_{\varepsilon}) \leq U(f_{N}, P_{\varepsilon}) + \frac{\varepsilon}{4}\]

since

\[U(f, P_{\varepsilon}) = \sum_{i=0}^{n-1} \sup\{ f(x):a_{i}\leq x\leq a_{i+1} \} (a_{i+1}-a_{i}).\]

In the same way, we have that

\[U(f_{N}, P_{\varepsilon}) - \frac{\varepsilon}{4}\leq U(f, P_{\varepsilon}),\]

so that

\[U(f_{N}, P_{\varepsilon}) - \frac{\varepsilon}{4} \leq U(f, P_{\varepsilon}) \leq U(f_{N}, P_{\varepsilon}) + \frac{\varepsilon}{4}.\]

In an analogous way (but working with infima instead of suprema) we can prove that

\[L(f_{N}, P_{\varepsilon}) - \frac{\varepsilon}{4}\leq L(f, P_{\varepsilon}) \leq L(f_{N}, P_{\varepsilon}) + \frac{\varepsilon}{4}.\]

Finally, we have that

\[\begin{aligned} U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) &\leq U(f_{N}, P_{\varepsilon}) + \frac{\varepsilon}{4} - \left( L(f_{N}, P_{\varepsilon}) - \frac{\varepsilon}{4} \right) \\ &= U(f_{N}, P_{\varepsilon}) - L(f_{N}, P_{\varepsilon}) + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \end{aligned}\]

from which we conclude that \(f\) is Riemann integrable. For the proof of the limit of the integrals: given \(\varepsilon>0\), there exists \(M \in \mathbb{N}\) such that for every \(n\geq M\) and \(x \in [a,b]\), \(\lvert f(x) - f_{n}(x) \rvert < \frac{\varepsilon}{b-a}\) Take

\[\begin{aligned} \left\lvert \int_{a}^{b} f_{n}(x) \, dx - \int_{a}^{b} f(x) \, dx \right\rvert &= \left\lvert \int_{a}^{b} (f_{n}(x) - f(x)) \, dx \right\rvert \\ &\leq \int_{a}^{b} \lvert f_{n}(x)-f(x) \rvert \, dx \\ &< \int_{a}^{b} \frac{\varepsilon}{b-a} \, dx = \varepsilon, \end{aligned}\]

whenever \(n\geq M\), from which the limit follows.

Example

For \(n \in \mathbb{N}\), consider \(f_{n}:[0,1] \to \mathbb{R}\) such that

\[f_{n}(x) = \begin{cases} 0 \quad\text{if } \frac{1}{n}\leq x\leq 1 \\ n^{2}x \quad \text{if } 0\leq x < \frac{1}{n} \end{cases}.\]

Let \(x_{0} \in (0,1)\). Then there exists \(n_{0}\) such that \(\frac{1}{n_{0}} \leq x_{0}\). Then \(f_{n}(x_{0}) = 0\) if \(n \geq n_{0}\). Now,

\[\int_{0}^{1} f_{n}(x) \, d = \int_{0}^{1/n} n ^{2}x\, dx + \int_{1/n}^{1} 0 \, dx = \frac{n^{2}x^{2}}{2} \biggr\rvert_{0}^{1/n} = \frac{1}{2}.\]

Then

\[\int_{0}^{1} f_{n}(x) \, dx = \frac{1}{2} \underset{n \rightarrow \infty}{ \not\longrightarrow} \int_{0}^{1} 0 \, dx = 0.\]

Exercise

Find a sequence of Riemann integrable functions \(f_{n}\) such that

\[f_{n} \underset{n \rightarrow \infty}{\longrightarrow} f = \begin{cases} 1 \quad \text{if }x \in \mathbb{Q} \\ \\ 0 \quad \text{if }x \in \mathbb{I} \end{cases},\]

with \(f_{n}:[0,1]\to \mathbb{R}\) and \(f : [0,1] \to \mathbb{R}\).

Theorem (Derivatives and uniform convergence)

Let \(f_{n}:[a,b]\to \mathbb{R}\) be differentiable functions such that \(f_{n}' \underset{n \rightarrow \infty}{\longrightarrow} g\) uniformly on \([a,b]\). Assume that there exists \(x_{0} \in [a,b]\) such that \(f_{n}(x_{0})\) converges. Then \(f_{n}\) converges uniformly to a function \(f:[a,b] \to \mathbb{R}\) such that \(f'(x) = g(x)\) for every \(x \in [a,b]\).

Proof: Let \(n,m \in \mathbb{N}\). Take \(h(x) = f_{n}(x)-f_{m}(x)\). Then, by the mean value theorem, there exists \(y\) between \(x\) and \(x_{0}\) such that

\[\begin{aligned} \frac{h(x) - h(x_{0})}{x-x_{0}} &= h'(y) \implies h(x) = h(x_{0}) + h'(y)(x-x_{0}) \\ \implies f_{n}(x) - f_{m}(x) &= (f_{n}(x_{0})-f_{m}(x_{0})) + (x-x_{0})(f_{n}'(y)-f_{m}'(y)) \end{aligned}\]

Since \(f_{n}' \underset{n \rightarrow \infty}{\longrightarrow} g\), it is Cauchy, i.e., given \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that for all \(n,m\geq N\) and every \(x \in [a,b]\), \(\lvert f_{n}'(x) - f_{m}'(x)\rvert < \frac{\varepsilon}{2(b-a)}\). Moreover, since \(f_{n}(x_{0})\) converges, it is Cauchy, so there exists \(M \in \mathbb{N}\) such that if \(n,m\geq M\), \(\lvert f_{n}(x_{0}) - f_{m}(x_{0}) \rvert < \frac{\varepsilon}{2}\). Hence, if \(n,m\geq N\) then, for every \(x \in [a,b]\)

\[\begin{aligned} \lvert f_{n}(x) - f_{m}(x) \rvert &\leq \lvert f_{n}(x_{0}) - f_{m}(x_{0}) \rvert + \lvert x-x_{0} \rvert \lvert f_{n}'(y) - f_{m}'(y) \rvert\\ &< \frac{\varepsilon}{2} + (b-a)\frac{\varepsilon}{2(b-a)} = \varepsilon. \end{aligned}\]

Therefore \(f_{n}\) is uniformly Cauchy and converges. CAREFUL: If \(\{f_{n}(x)\}_{n=1}^\infty\) is Cauchy, define \(\ell_{x} = \lim_{ n \to \infty } f_{n}(x)\). Then \(f(x) = \ell_{x}\).

We will now prove that \(f'(x) = g(x)\). We must show that

\[\lim_{ x \to y } \frac{f(x)-f(y)}{x-y} = g(y).\]

Consider

\[\begin{aligned} &\left\lvert \frac{f(x)-f(y)}{x-y} - g(y) \right\rvert \leq \left\lvert \frac{f(x)-f(y)}{x-y} -\frac{f_{N}(x)-f_{N}(y)}{x-y} \right\rvert + \left\lvert \frac{f_{N}(x)-f_{N}(y)}{x-y} - g(y) \right\rvert \\ &\leq \left\lvert \frac{f(x)-f(y)}{x-y} -\frac{f_{N}(x)-f_{N}(y)}{x-y} \right\rvert + \left\lvert \frac{f_{N}(x)-f_{N}(y)}{x-y} - f_{N}'(y) \right\rvert + \lvert f_{N}'(y) - g(y) \rvert . \end{aligned}\]

Given \(\varepsilon>0\), there exist \(N \in \mathbb{N}\) and \(\delta>0\) such that

1. \(\lvert f_{n}'(x) - g(x) \rvert < \frac{\varepsilon}{3}\) if \(n\geq N\) and \(x \in [a,b]\).
2. If \(\lvert x-y \rvert < \delta\) then

\[\left\lvert \frac{f_{N}(x)-f_{N}(y)}{x-y} - f_{N}'(y) \right\rvert < \frac{\varepsilon}{3}.\]

Now consider

\[\begin{aligned} \frac{f_{m}(x)-f_{m}(y)}{x-y} - \frac{f_{N}(x)-f_{N}(y)}{x-y} &= \frac{(f_{m}(x) - f_{N}(x))- (f_{m}(y)-f_{N}(y))}{x-y} \\ &= f_{m}'(c)-f_{N}'(c) \end{aligned}\]

for some \(c\) between \(x\) and \(y\) (which exists by the intermediate value theorem). Then

\[\lim_{ m \to \infty } \left\lvert \frac{f_{m}(x)-f_{m}(y)}{x-y} - \frac{f_{N}(x)-f_{N}(y)}{x-y}\right\rvert = \left\lvert \frac{f(x)-f(y)}{x-y} - \frac{f_{N}(x)-f_{N}(y)}{x-y} \right\rvert < \frac{\varepsilon}{3}.\]

Finally, we have that

\[\left\lvert \frac{f(x)-f(y)}{x-y} - g(y) \right\rvert < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.\]

Example

Let \(f_{n}:[-1,1] \to \mathbb{R}\) be such that \(f_{n}(x) = \sqrt{ x^{2}+\frac{1}{n} }\). Then \(f_{n}\) is differentiable on \((0,1)\) and \(f_{n}'(x) = \frac{x}{\sqrt{ x^{2}+\frac{1}{n} }}\). Moreover \(\lim_{ n \to \infty } f_{n}(x) = \sqrt{ x^{2} } = \lvert x \rvert\). We are going to prove that, given \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that \(\left\lvert \sqrt{ x^{2}+\frac{1}{n} } - \sqrt{ x^{2} } \right\rvert < \varepsilon\) for every \(x \in [-1,1]\). Given \(\varepsilon>0\), we know that there exists \(N_{0} \in \mathbb{N}\) such that \(\frac{1}{\sqrt{ N_{0}}} < \varepsilon\). Note that, for \(n\geq N_{0}\)

\[\sqrt{ x^{2}+\frac{1}{n} } - \sqrt{ x^{2} } = \frac{\frac{1}{n}}{\sqrt{ x^{2}+\frac{1}{n} } + \sqrt{ x^{2} }} \leq \frac{\frac{1}{n}}{\sqrt{ \frac{1}{n} }} = \frac{1}{\sqrt{ n }} < \varepsilon\]

Example

\[f_{n}(x) = \sum_{k=0}^{n} \frac{\cos(3^{k}x)}{2^{k}}\]

converges uniformly to a function that is nowhere differentiable.

Lemma (Pointwise convergence of increasing functions)

Let \(\{f_{n}(x)\}_{n=1}^\infty\) be continuous with \(f_{n}(x) \leq f_{n+1}(x)\) for every \(x \in [a,b]\). Assume that there exists a continuous \(f:[a,b] \to \mathbb{R}\) such that \(\lim_{ n \to \infty }f_{n}(x) = f(x)\) pointwise. Then \(f_{n} \underset{n \rightarrow \infty}{\longrightarrow} f\) uniformly on \([a,b]\).

Proof: We must show that \(M_{n} = \lVert f-f_{n} \rVert_{\infty} \underset{n \rightarrow \infty}{\longrightarrow} 0\), where \(\lVert f - f_{n} \rVert_{\infty} = \sup \{ \lvert f(x) -f_{n}(x) \rvert: x \in A \}\). Note that for every \(x \in A\),

\[0 \leq f(x) - f_{n+1}(x) \leq f(x)-f_{n}(x) \implies M_{n+1} \leq M_{n}.\]

Since \(f - f_{n}\) is continuous, we know that there exists \(x_{n}\) such that \(f(x_{n}) - f_{n}(x_{n}) = M_{n}\). In particular, \(\{x_{n}\}_{n=1}^\infty \subseteq [a,b]\). By Bolzano-Weierstrass, there exists a subsequence \(x_{n_{k}} \underset{n_{k} \rightarrow \infty}{\longrightarrow} x \in[a,b]\). Moreover, note that \(f_{1}(x_{n_{k}}) \leq f_{{n_{k}}}(x_{n_{k}}) \leq f(x_{n_{k}})\). Then, by BW, there exists a subsequence \(x_{n_{k_{\ell}}}\) such that \(f_{n_{k_{\ell}}}(x_{n_{k_{\ell}}})\) is convergent, (\(f_{n}(x_{n})\)). Without loss of generality I may assume that …