What properties of functions are preserved under pointwise limits? Let $f_n \to f$ pointwise at every point in the interval $[a,b]$. And $f_n$ is continuous . Which of the following are true ?

Let $f_n \to f$ pointwise at every point in the interval $[a,b]$.  We have seen that even if each $f_n$ is continuous it does not follow that $f$ is continuous.  Which of the following statements are true?

*

*If each $f_n$ is increasing on $[a,b]$, then so is $f$.

*If each $f_n$ is nondecreasing on $[a,b]$, then so is $f$.

*If each $f_n$ is bounded on $[a,b]$, then so is $f$.

*If each $f_n$ is everywhere discontinuous on $[a,b]$, then so is $f$.

*If each $f_n$ is constant on $[a,b]$, then so is $f$.

*If each $f_n$ is positive on $[a,b]$, then so is $f$.

*If each $f_n$ is linear on $[a,b]$, then so is $f$.


I guess  (1) and (3) are true but I am not sure.
 A: *

*False.  Consider the sequence $f_n : [0,1]\to\mathbb{R}$ defined by $f_n(x) = x^n$.  The functions $f_n$ are strictly increasing for each $n$, but the limiting function is $\chi_{\{1\}}$, which is zero on $[0,1)$, and therefore not increasing on $[0,1]$.

*True.  The argument is a little tricky.  Fix $x<y\in[a,b]$, suppose for contradiction that $f(x) > f(y)$, and let $\varepsilon = f(x) - f(y)$.  By pointwise convergence, there is some $N$ sufficiently large that $n \ge N$ implies that
$$ |f_n(x) - f(x)| < \frac{\varepsilon}{3}
\qquad\text{and}\qquad
|f_n(y) - f(y)| < \frac{\varepsilon}{3}. $$
But then
\begin{align} &-\frac{2\varepsilon}{3} < (f_n(x) - f(x)) - (f_n(y) - f(y)) < \frac{ 2\varepsilon}{3} \\
&\qquad\qquad\implies (f(x)-f(y)) - \frac{2\varepsilon}{3} = \frac{\varepsilon}{3} < f_n(x)-f_n(y),\end{align}
which contradicts the hypothesis that $f_n$ is nondecreasing.

*False.  Consider the sequence $f_n : [0,1] \to \mathbb{R}$.
$$ f_n(x) = \begin{cases}
0 & \text{if $x\in[0,\frac{1}{n})$} \\
\frac{1}{x} & \text{otherwise.} \\
\end{cases} $$

*False.  Consider $f_n = \frac{1}{n} \chi_{\mathbb{Q}}$, restricted to any interval you like.

*True.  If $x,y\in [a,b]$, then we can choose $N$ so large that $n\ge N$ implies that
\begin{align}
|f(x) - f(y)| 
&= |f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) | \\
&\le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y) | \\
&\le \frac{\varepsilon}{2} + 0 + \frac{\varepsilon}{2} \\
&= \varepsilon.
\end{align}
Therefore $|f(x) - f(y)| = 0$, i.e. $f(x) = f(y)$.
Alternatively, as Kavi Rama Murthy suggests, fix some $x_0\in [a,b]$.  Then by the definition of $f$, we have
$$ f(x_0) = \lim_{n\to \infty} f_n(x_0) = \lim_{n\to \infty} c_n, $$
where $f_n(x) = c_n$ for all $x\in [a,b]$.  But then for any $x\in [a,b]$, we have
$$ f(x) = \lim_{n\to \infty} f_n(x_0) = \lim_{n\to \infty} c_n = f(x_0), $$
which implies that $f$ is constant.

*False.  Consider $f_n(x) = \frac{1}{n}$, which tends pointswise to 0, which is not positive.  Replace "positive" with "nonnegative" to get something that is true (do you see why?).

*And I'll leave you one to work on by yourself.  This one shouldn't be too hard. ;)

