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(Hint: note that \(\widetilde{x}\) is a simple zero implies that \(f^{\prime}(x)\neq 0\) on some neighborhood \(U\) of \(\widetilde{x}\) where \(U\subset[a,b]\) and \(f^{\prime\prime}\) is bounded on \(U\)).
11. Show that the Fredholm integral equation \(y(x)=1+\lambda\int_{0}^{1}e^{x-t}y(t)\,dt\) has a solution for \(|\lambda|<\frac{1}{e}\). Set up an iteration scheme to show that
\[\lim_{n\to\infty}y_{n}(t)=1-\frac{\lambda}{1-\lambda}e^{x}\left(\frac{1}{e}- 1\right).\]
12. Let \(k(x,y)\) be a continuous function on the unit square \(0\leq x,y\leq 1\) satisfying \(|k(x,y)|<1\) for all \(x\) and \(y\). Show that there is a continuous function \(f(x)\) on \([0,1]\) such that one has
\[e^{x^{2}}=f(x)+\int_{0}^{1}k(x,y)f(y)dy.\]
Can there be more than one such function \(f\)?
### 2.3 Modes of Convergence
#### Pointwise and Uniform Convergence
Suppose that we are given a sequence of functions \(f_{1},f_{2},\dots\) all defined on a subset \(D\) of \(\mathbb{R}\). For each fixed point \(x_{0}\in D\) we form the numerical sequence \(\{f_{n}(x_{0})\}\). If \(\{f_{n}(x_{0})\}\) converges to \(f(x_{0})\) as sequence of real numbers for every \(x_{0}\in D\), we say \(\{f_{n}\}\)_converges pointwise_ to \(f\) and write \(f_{n}\to f\). We will soon see that desirable properties of the functions \(f_{n}\), such as continuity and integrability, need not be preserved under pointwise convergence. However, under a stronger type of convergence known as _uniform convergence_, the situation is much more positive.
**Definition 74**.: A sequence of functions \(\{f_{n}\}\) is _uniformly convergent_ to \(f\) on the set \(D\) if for each \(\epsilon>0\) there exists a number \(N\) such that \(|f_{n}(x)-f(x)|<\epsilon\) for all \(x\in D\) whenever \(n\geq N\). We use \(f_{n}\rightrightarrows f\) to describe uniform convergence (Figure 2.12).
Note that uniform convergence does depend on the set \(D\) being considered. Convergence could be uniform in one domain, but not on a larger domain. Clearly uniform convergence implies pointwise convergence; however, it imposes a stronger requirement on the number \(N\) so that \(N\) depends only on \(\epsilon\) and not on the particular point \(x_{0}\in D\).
**Example 81**.: The following are two examples of pointwise convergence.
* The functions \(f_{n}(x)=\dfrac{x}{n}\) converge pointwise to \(f(x)\equiv 0\). Clearly these functions are just lines with decreasing slopes, but the convergence is not uniform on \(\mathbb{R}\).
* The functions \(f_{n}(x)=x^{n}\) converge pointwise but not uniformly in the interval \([0,1]\) to the limit \(f(x)=0\) for \(x\in[0,1)\) and \(f(1)=1\). Note that each of the functions \(x^{n}\) is continuous; however, their limit \(f\) is not (Figure 2.13). Furthermore, for \(x>1\), \(\lim\limits_{n\to\infty}x^{n}\) does not exist and therefore for \(x>1\)\(\{f_{n}\}\) does not converge pointwise. However, if we restrict the domain to \([0,1]\), then \[\lim\limits_{n\to\infty}x^{n}=\begin{cases}0&\text{if}\ \ x\in[0,1)\\ 1&\text{if}\ \ x=1\end{cases}.\]So we have pointwise convergence to \(f\), however, \(f\) is not continuous. The following theorem illustrates that continuity is preserved under uniform convergence.
**Theorem 90** (Interchange of limit operations).: _Suppose that \(f_{n}\) for \(n=1,2,\dots\), are defined and continuous on \(D\subset\mathbb{R}\) such that \(f_{n}\rightrightarrows f\) (uniformly) as \(n\to\infty\). Then \(f\) is continuous on \(D\)._
Proof
: Let \(\epsilon>0\) be given and let \(x_{0}\) be any point in \(D\). From
\[|f(x)-f(x_{0})|=|f(x)-f_{n}(x)+f_{n}(x)-f_{n}(x_{0})+f_{n}(x_{0})-f(x_{0})|,\]
the inequality
\[|f(x)-f(x_{0})|\leq|f(x)-f_{n}(x)|+|f_{n}(x)-f_{n}(x_{0})|+|f_{n}(x_{0})-f(x_{0})|\]
holds for all \(x\in D\). Because \(f_{n}\rightrightarrows f\), there exists a positive integer \(N\) depending only on \(\epsilon\) such that \(|f(x)-f_{n}(x)|\leq\frac{\epsilon}{3}\) and \(|f_{n}(x_{0})-f(x_{0})|\leq\frac{\epsilon}{3}\) for all \(x\in D\) whenever \(n\geq N\). Choose \(n\geq N\) and use the fact that \(f_{n}\) is continuous at \(x_{0}\) to conclude \(|f_{n}(x)-f_{n}(x_{0})|\leq\frac{\epsilon}{3}\) holds. Therefore
\[|f(x)-f(x_{0})|\leq\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\]
holds for all \(x\in D\) with \(|x-x_{0}|<\delta\). Thus \(f\) is continuous at \(x_{0}\).
Note that under conditions of the above theorem, when \(\{x_{m}\}\) is any convergent sequence in \(D\) whose limit \(x\) is in \(D\), then by the continuity of \(f_{n}\)’s we have \(\lim\limits_{m\to\infty}f_{n}(x_{m})=f_{n}(x)\), therefore
\[\lim\limits_{n\to\infty}(\lim\limits_{m\to\infty}f_{n}(x_{m}))=\lim\limits_{ m\to\infty}(\lim\limits_{n\to\infty}f_{n}(x_{m})).\]
Thus we see that limit operations can be interchanged.
Figure 2.13: Limit function is not continuous
**Definition 75**.: Let \(C[a,b]\) denote the space of all continuous real-valued functions defined on \([a,b]\). For \(f\in C[a,b]\), define
\[||f||=\sup\{|f(x)|:\ x\in[a,b]\}.\]
We say \(f_{n}\) converges to \(f\) in \(C[a,b]\) if and only if \(||f_{n}-f||\to 0\).
**Theorem 91** (Uniform convergence in the \(\sup\) norm).: \(\{f_{n}\}\) _converges uniformly to \(f\) on \([a,b]\) if and only if \(\{f_{n}\}\) converges to \(f\) in \(C[a,b]\)._
Proof
: Follows directly from definitions.
### Test for Uniform Convergence
**Theorem 92** (Test for Uniform Convergence of Sequences).: _A sequence of functions with domain \(D\) converges uniformly to a function \(f\) if there is a real-valued sequence \(\{a_{n}\}\) such that \(a_{n}\to 0\) and_
\[|f_{n}(x)-f(x)|\leq|a_{n}|\]
_for all \(x\in D\) and for all \(n\in N\)._
Proof
: We know that \(a_{n}\to 0\). So, given \(\epsilon>0\) there exists \(n\in\mathbb{N}\) such that if \(n>N\), then \(|a_{n}|<\epsilon\). This \(N\) is independent of \(x\in D\) and so
\[|f_{n}(x)-f(x)|\leq|a_{n}|<\epsilon\]
for all \(x\in D\) whenever \(n>N\). This implies that \(\{f_{n}\}\) converges uniformly.
**Theorem 93**.: _Let \(\{f_{n}\}\) be a sequence of functions defined on \(A\). Then \(f_{n}\rightrightarrows f\) uniformly on \(B\subset A\)\(\Leftrightarrow\) the sequence of numbers_
\[d_{n}:=\sup\{|f_{n}(x)-f(x)|:x\in B,n\in\mathbb{N}\}\]
_converges to \(0\)._
Proof
: \(\Rightarrow\) Suppose that \(f_{n}\rightrightarrows f\) uniformly. Then, given \(\epsilon>0\) there exists \(N\in\mathbb{N}\) such that \(|f_{n}(x)-f(x)|<\epsilon\) for all \(n>N\) and for all \(x\in B\). So, for \(n>N\), \(d_{n}<\epsilon\). Therefore, \(d_{n}\to 0\).
\(\Leftarrow\) Suppose \(d_{n}\to 0\). Then we know that for sufficiently large \(n\) and for all \(x\in B\),
\[|f_{n}(x)-f(x)| So \(f_{n}\rightrightarrows f\) uniformly. **Remark 47**.: Note that Theorems 91, 92, and 93 are several interpretations of the fact that \[f_{n}\rightrightarrows f\quad\text{uniformly}\quad\text{ if and only if}\quad||f_{n}-f||\to 0.\] For example for Theorem 92 we can just take \(a_{n}=||f_{n}-f||\). **Example 82**.: Let \(f_{n}(x)=\dfrac{nx^{2}}{1+nx}\) and \(A=[0,1]\). We claim that \(\{f_{n}(x)\}\) converges uniformly. Note that \(\lim_{n\to\infty}f_{n}(x)=\lim_{n\to\infty}\dfrac{nx^{2}}{1+nx}=x.\) So, we have \[|f_{n}(x)-f(x)|=\left|\dfrac{nx^{2}}{1+nx}-x\right|=\left|\dfrac{x}{1+nx} \right|.\]
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