Answer to Question #347914 in Real Analysis for Nikhil

Question

The sum of two discontinuous function is always discontinuous function. True or false with full explanation

Answer 1

ANSWER . The statement is false.

EXPLANATION

The sum of two discontinuous functions can be a continuous function (example 1), can be a discontinuous function (example 2)

Example 1

$h(x) = \begin{cases}1 & \text{ if } -2\leq x\leq 0 \\ -1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}\\g(x) = \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3}+1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}$

These function are discontinuous at $x=0$ .

$f(x)=h(x)+g(x)=x^{3}, x\in[-2,2]$ is continuous .

Example 2

$h(x) = \begin{cases}0 & \text{ if } -2\leq x\leq 0 \\ -1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}\\g(x) = \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3}+1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}$

These function are discontinuous at $x=0$ .

$f(x) =h(x)+g(x)= \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3} & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}$

This function is discontinuous at x=0