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Suppose $f$ is an increasing real-valued function on [$0, ∞$) with $f(x)> 0$ for all $x$ and let
$g (x) = (1/x)∫_0^xf(u)du$
Then which of the following are true:
1. $g(x) ≤ f(x)$ for all $x∈(0,∞)$
2. $xg(x) ≤ f(x)$ for all $x∈ (0, ∞)$
3. $xg(x) ≥ f(0)$ for all $x∈ (0, ∞)$
4. $yg(y) – xg(x) ≤ (y-x)f(y)$ for all $x < y$.

how should i solve this problem.i am completely stuck on it.

2 Answers 2

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4.

$y g(y) - x g(x) = \int_x^y du f(u) \le (y-x) f(y)$

because $f$ is increasing.

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  1. g is the average of f. Since f is positive and increasing, its average is less than its value. Explicitly, $f(x) - g(x) = f(x) - (1/x)∫_0^xf(u)du = (1/x)∫_0^x(f(x)-f(u))du \ge 0$ since $f(x) \ge f(u) $ .

  2. This can be true or false depending on how fast f grows. If $f(x) = x^a+b$, $xg(x) = ∫_0^xf(u)du = ∫_0^x(u^a+b)du = x^{a+1}/(a+1) + bx $. If $a=1$, $xg(x)=x^2/2+bx$, so $xg(x) < f(x)$ if $x$ is small and $b < 1$ and $xg(x) > f(x)$ if $b > 1$ and $x$ is large.

  3. This is false. Look at what happens at $x = 0$.