MHB Is non continuous function also not Bounded ?

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A non-continuous function can still be bounded, as demonstrated by examples of functions that are discontinuous yet remain within a finite range. The logic that a non-continuous function must be unbounded is flawed; the relationship between continuity and boundedness does not imply that non-continuity leads to unboundedness. Continuous functions in a closed interval are indeed bounded, but the reverse is not true for non-continuous functions. Therefore, it is possible for non-continuous functions to be bounded. Understanding this distinction is crucial in mathematical analysis.
Lancelot1
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Dear all,

I am trying to figure out if a non continuous function is also not bounded. I know that a continuous function in an interval, closed interval, is also bounded. Is a non continuous function in a closed interval not bounded ? I think not, it makes no sense. How do you prove it ?

Thank you !
 
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You are making an error in logic. The opposite of "if A then B" is NOT "if not A then not B". It is "if not A then we don't know anything about B"!

For example the functions f(x)= x for [math]0\le x\le 1[/math], f(x)= x+ 1 for [math]1\le x\le 2[/math] and g(x)= 0 if x is rational, g(x)= 1 if x is irrational are discontinuous (g badly discontinuous) but are bounded functions.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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