What properties of functions are preserved under pointwise limits?

Question

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?

1. If each $f_n$ is increasing on $[a,b]$, then so is $f$.
2. If each $f_n$ is nondecreasing on $[a,b]$, then so is $f$.
3. If each $f_n$ is bounded on $[a,b]$, then so is $f$.
4. If each $f_n$ is everywhere discontinuous on $[a,b]$, then so is $f$.
5. If each $f_n$ is constant on $[a,b]$, then so is $f$.
6. If each $f_n$ is positive on $[a,b]$, then so is $f$.
7. 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.

Answer 1

1. 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 $[0,1]$.
2. 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.
3. False. Consider the sequence $f_n : [0,1] \to \mathbb{R}$. $$ f_n(x) = \begin{cases} 0 & \text{if $x\in[0,\frac{1}{n})$} \\ \frac{1}{x} & \text{otherwise.} \\ \end{cases} $$
4. False. Consider $f_n = \frac{1}{n} \chi_{\mathbb{Q}}$, restricted to any interval you like.

5. 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.

6. 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?).
7. And I'll leave you one to work on by yourself. This one shouldn't be too hard. ;)