My question: Suppose that the sequence of functions $(f_n)$ converges pointwise on $A$ to both functions $f$ and $g$. Show that $f = g$, i.e. the limit is unique.
I'm not really sure how to start, but do I need to show this by using the definition of pointwise convergence? Or can I do this via contradiction?
Many thanks!
On
For fixed $x$, let $\epsilon>0$, find some $N$ (depends on $x$, $\epsilon$, and the functions $f,g$) such that $|f_{N}(x)-f(x)|<\epsilon/2$ and $|f_{N}(x)-g(x)|<\epsilon/2$, then $|f(x)-g(x)|\leq|f_{N}(x)-f(x)|+|f_{N}(x)-g(x)|<\epsilon$, this is true for all $\epsilon>0$, so $|f(x)-g(x)|=0$, or $f(x)=g(x)$, this is true for all $x$, so $f=g$.
According to the definition of point-wise convergence, since $\lim_{n\to\infty}f_n=f$ we have
\begin{align*} \forall x\,\,\forall\epsilon_1\gt0\,\,\exists N_1\in\mathbb{Z}^+\,\,:n\ge N_1 \implies |f_n(x)-f(x)|\le\epsilon_1 \end{align*}
where $N_1$ may depend on $x$ and $\epsilon_1$. Also $\lim_{n\to\infty}f_n=g$ and we we have
\begin{align*} \forall x\,\,\forall\epsilon_2\gt0\,\,\exists N_2\in\mathbb{Z}^+\,\,:n\ge N_2 \implies |f_n(x)-g(x)|\le\epsilon_2 \end{align*}
where $N_2$ may depend on $x$ and $\epsilon_2$. Let $N=\max\{N_1,N_2\}$ so both of the above inequalities hold for $n\ge N$. Given $\epsilon\gt 0$, choose $\epsilon_1=\epsilon_2=\frac{\epsilon}{2}$. As we will see, any other choice which makes $\epsilon_1+\epsilon_2\le \epsilon$ will work. Now using the triangle inequality we have
\begin{align*} |f(x)-g(x)|\le|f(x)-f_n(x)|+|f_n(x)-g(x)|\lt\epsilon_1+\epsilon_2=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \end{align*}
for every $\epsilon>0$ and $n\ge N$. But according to the epsilon principle this means that $f(x)=g(x)$ for all $x$ which means $f=g$.