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Let, f and g be two functions on R, support IVP and at least one them is discontinuous. Then prove or disprove ( with example) whether f+g also supports IVP.

If f and g, both are continuous, then it is easy to say that, f+g supports intermediate-value-property. But here, we have to show, whether this result holds for discontinuous function or not.

Amlan Saha Kundu
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  • I found a related answer: https://math.stackexchange.com/a/624120/58577 This shows that every function is the sum of two functions satisfying the IVP. – gerw Jan 17 '19 at 18:11

2 Answers2

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Consider the interval $I=[0,1]$.

$f(x)=x$ is continuous on $I$. The map $$h(x)=\begin{cases} 0 & \mbox{for} &0 \le x <1/2\\ 1 & \mbox{for} &1/2 \le x \le 1 \end{cases}$$

Doesn't support the IVP. However $g=h-f$ which is discontinuous does.

Unfortunately my answer is wrong...

mathcounterexamples.net
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  • Thanks for your answer. But here, we have to prove whether, summation of two functions that support ivp (at least one discontinuous) supports ivp. Since h(x) is discontinuous but doesn't support IVP, this example is not valid. – Amlan Saha Kundu Jan 17 '19 at 14:24
  • Please read again. $f$ is continuous. $g$ is discontinuous and supports IVP. $f+g=h$ doesn't support IVP. This answers your question (negatively). – mathcounterexamples.net Jan 17 '19 at 14:26
  • I have tried with f as x and g as - {x}, and come up with [x] which doesn't support IVP. Since, {x} doesn't support IVP, my example was not valid. – Amlan Saha Kundu Jan 17 '19 at 14:26
  • I am extremely sorry. Yes, it does. Thanks a lot. – Amlan Saha Kundu Jan 17 '19 at 14:29
  • I am confused. Why does $g$ satisfies IVP? In fact, $g(1/4) = -1/4$, $g(3/4) = 1/4$, but $g(s) < -1/4$ for $s \in (1/4,1/2)$ and $g(s) > 1/4$ for $s \in [1/2,3/4)$. In particular, there is no $s \in (1/4,3/4)$ with $g(s) = 0$. Where is my mistake? – gerw Jan 17 '19 at 14:44
  • @gerw That is good point. I'll delete the answer after its rejection. – mathcounterexamples.net Jan 17 '19 at 14:53
  • You might want to check my (counter)example – gerw Jan 17 '19 at 14:57
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For any $s \in [-1,1]$, one can show that $f_s : \mathbb R \to \mathbb R$ defined via $$f_s(x):= \begin{cases} \sin(1/x) & \text{if } x \ne 0 \\ s & \text{else}\end{cases}$$ satisfies the IVP.

However, the difference of two such functions does clearly violate the IVP.

gerw
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