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Let $f$ and $g$ be continuous functions such that $f(x) ≤ g(x)$ for all $x ∈ [0, 1]$. Determine which of the following statements are true and which are false:

$$ \begin{align} (a) & {}\quad \int_0^x |f(t)|~dt \leq\int_0^x |g(t)|~dt ~\forall~x ∈ [0, 1]\\ (b) & {}\quad \int_0^x (|f(t)|+f(t))~dt \leq\int_0^x \left(|g(t)|+g(t)\right)~dt ~\forall~x ∈ [0, 1]\\ (c) & {}\quad \int_0^x (|f(t)|-f(t))~dt \leq\int_0^x \left(|g(t)|-g(t)\right)~dt ~\forall~x ∈ [0, 1] \end{align}$$ For any statement which you believe to be true, you need to give a proof and for any statement which you believe to be false, you need to give a counter example.

Where to start and how I proceed? Please help me.

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I am stuck in first step.Please give me some clue so that I can proceed – Argha Aug 5 '12 at 4:45
@Ranabir I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware about the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more question, stackexchange software will not allow you to do so.) For more details see meta. – Martin Sleziak Aug 5 '12 at 8:05
up vote 2 down vote accepted

Some hints.

a) Consider $f$ negative and $g$ positive with $f$ having a larger absolute value at all points.

b) Note that $|f(x)| + f(x)$ is $2f(x)$ for $f(x) \ge 0$ and $0$ for $f(x) < 0$. So consider the the sign of the functions. When $f$ is positive and certainly $g$ is as well, what can you say about the integrands? If $f(x)$ is negative, then |f(x)| + f(x) is $0$, what can you say about the integrand of $|g(x)| + g(x)$?

c) Idea is similar to b.

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Hint for $(a)$: Take $f(x)\equiv -1$ and $x=1$, can you think of a 'simple' $g(x)$ to give a counter-example ?

I hope that this will get you started on $(b),(c)$ as well

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You should be able to find counterexamples for (a) and (c) reasonably easily. If you consider constant functions, this will make the integration easier.

For (b), you'll have to break it down into steps. You could start by thinking about whether or not $f(x)\leq g(x)$ implies that $|f(x)|+f(x) \leq |g(x)|+g(x)$, at a particular value of $x$. Consider three cases, depending on whether or not each of $f(x), g(x)$ is $\geq 0$. Once you have an understanding of whether this implication holds, then consider what happens when you do the integration.

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(a)Let $f(x)=-2~\forall~x\in[0,1]$ and $g(x)=1~\forall~x\in[0,1]$.Here $f(x)\leq g(x)~\forall~x\in[0,1]$ and also $2\int_0^x ~dt \not\leq\int_0^x~dt ~\forall~x ∈ [0, 1]$ .So the statement is false

(b)Here we need to consider three cases.

Case 1:$f(x)<0$ and $g(x)<0~\forall~x\in[0,1]$ $$=>|f(t)|+f(t)=0,|g(t)|+g(t)=0$$$$=>\int_0^x (|f(t)|+f(t))~dt=\int_0^x (|g(t)|+g(t))~dt~\forall~x\in[0,1]$$Case 2:$f(x)<0$ and $g(x)\geq0~\forall~x\in[0,1]$$$=>|f(t)|+f(t)=0,|g(t)|+g(t)=2g(t)$$$$=>\int_0^x (|f(t)|+f(t))~dt\leq\int_0^x (|g(t)|+g(t))~dt~\forall~x\in[0,1]$$Case 3:$f(x)\geq0$ and $g(x)\geq0~\forall~x\in[0,1]$$$=>|f(t)|+f(t)=2f(t),|g(t)|+g(t)=2g(t)$$$$=>\int_0^x (|f(t)|+f(t))~dt\leq\int_0^x (|g(t)|+g(t))~dt~\forall~x\in[0,1]$$ Hence this statement is true.

(c)Let $f(x)=-2~\forall~x\in[0,1]$ and $g(x)=-1~\forall~x\in[0,1]$$$=>|f(t)|-f(t)=4,|g(t)|-g(t)=2$$$$=>\int_0^x (|f(t)|-f(t))~dt\not\leq\int_0^x (|g(t)|-g(t))~dt~\forall~x\in[0,1]$$Hence this statement is also false.

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I tried as you guide.Am I not make any mistake in any step?Can this procedure more improved? – Argha Aug 5 '12 at 10:40

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