1
$\begingroup$

Let $f(x)$ define a continuous function on some interval $\Bbb X$. Prove that the following functions are continuous in $\Bbb X$: $$ f_+(x) = \begin{cases} \begin{align} f(x),\ &f(x) > 0 \\ 0,\ &f(x) \le 0 \end{align} \end{cases} $$ $$ f_-(x) = \begin{cases} \begin{align} 0,\ &f(x) \ge 0 \\ f(x),\ &f(x) < 0 \end{align} \end{cases} $$

It looks like if we sum up both functions we eventually get $f(x)$ itself, so: $$ f(x) = f_-(x) + f_+(x) $$

My main idea was to arrive at a contradicton. The function is either continuous or discontinuous so we have $4$ possible cases:

  1. Both $f_-(x)$ and $f_+(x)$ are continuous
  2. Both $f_-(x)$ and $f_+(x)$ are discontinuous
  3. $f_-(x)$ is continuous and $f_+(x)$ is discontinuous
  4. $f_-(x)$ is discontinuous and $f_+(x)$ is continuous

I've proven a while ago that if a function $f$ is continuous and $g$ is discontinuous at some point then $f+g$ is also discontinuous at that point. That means for cases $3$ and $4$: $$ \begin{align*} f(x) = f_-(x) + f_+(x) \iff f(x) - f_-(x) = f_+(x)\tag 3\\ f(x) = f_-(x) + f_+(x) \iff f(x) - f_+(x) = f_-(x)\tag 4 \end{align*} $$ So we have arrived at a contradiction by the sum of continuous function must be continuous. By this we're left with cases $1$ and $2$.

Unfortunately, I couldn't eliminate case $2$ because the sum of discontinuous functions may be either continuous or discontinuous.

Is it possible to apply similar reasoning to eliminate case $2$? If not how do I show what's asked in the problem statement?

$\endgroup$
2
  • 2
    $\begingroup$ Can you use the fact that the composition of continuous functions is continuous? $\endgroup$ – eranreches Jun 17 '19 at 17:08
  • $\begingroup$ @eranreches Yes, I've proven that recently $\endgroup$ – roman Jun 17 '19 at 17:09
2
$\begingroup$

Hint: Prove that $g\left(x\right)=\max\left\{0,x\right\}$ (and similarly $\min$) is continuous. This is much simpler. Now use composition.

$\endgroup$
1
  • $\begingroup$ What a neat observation. Thank you! $\endgroup$ – roman Jun 17 '19 at 17:21
1
$\begingroup$

Note that $$f_+(x)=\max\{0,f(x)\}=\frac12 (f(x)+|f(x)|),$$ which is continuous as a sum of continuous functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.