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I posted this earlier and there is a question after this where I give an example of a function so that the inequality is strict. Basically my understanding is that they're looking for a function that is negative in a subinterval.

So, $f(x)=x^2-1$ in the interval $[-2,2]$ would have a strict inequality: $$\left|\int_a^bf(x)dx\right|<\int_a^b|f(x)|dx$$

Does this work? Does anybody have a "better" example that could enhance my understanding of this question perhaps? :) I'm just solving these to review/prepare for a midterm anyways

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just take $f(x) = -1$ on $[-1,0]$ and $f=1$ on $(0,1]$ ! if you're looking for a continuous example (which is not necessary for the inequality to hold) then yours is just fine.

note that this is the continuous analogue of the triangular inequality $|a+b| \leq |a| + |b|$

EDIT : if you are looking for an example which is true for ALL a and b then you'll have to work with some pretty complicated function. Let's work with a function defined on [0,1] (you can extend it to $R$ by periodicity if you want to).

Choose a Cantor set $C$ which has positive but not full measure on [0,1] (look up Cantor space if you don't know what it is : roughly speaking, it is obtained by removing an interval centered at the middle of [0,1] from [0,1], then removing another interval centered at the middle of those 2 intervals, etc. and by choosing carefully the lengths of the interval removed). Set $f(x) = 1$ if $x \in C$ and -1 otherwise. ($f$ is horrible)

Then for any $a$, $b$ in [0,1 the inequality will be strict as there will be a subset X of [a,b] such that $\int_X f <0$ and $\int_{[a,b]\backslash X} f>0$ wich is essentially what you need (compensation in the left term of the inequality).

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He is looking for an example which necessarily has the strict inequality. Your example doesn't seem to work for $a=0$ and $b=1$. –  process91 Nov 22 '11 at 23:55
    
@Glougloubarbaki Yes, you are right f simply has to be integrable, or almost continuous on an interval ;). I'm a little unsure if the function you put up fits this though... –  MathMathCookie Nov 22 '11 at 23:56
    
@process91 Yes, but it seems Glougloubarbaki set a=-1 and b=1. Or actually, the question verbatim states, "give an example where the inequality is strict." So would this mean that I don't have a choice of setting a,b equal to something to show an example? –  MathMathCookie Nov 22 '11 at 23:58
    
@MathMathCookie Perhaps so. I was operating under the assumption that it had to be true for all $a\ne b$, but that could be incorrect. –  process91 Nov 23 '11 at 0:00
    
well the inequality is strict iff $f$ has non constant sign on [a,b]. so if you require that it is strict for ALL a,b (which I didn't think was the case, as said @MathMathCookie), you won't find any continuous example. and for an example of that you'll need a Cantor... I'll edit my post. –  Glougloubarbaki Nov 23 '11 at 10:08

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