Answer to Question #347914 in Real Analysis for Nikhil

Question #347914

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

1
Expert's answer
2022-06-06T12:29:31-0400

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

These function are discontinuous at "x=0" .

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

This function is discontinuous at x=0

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