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Let $f(x)$ defined on $[0,a]$ with the following property.

  1. $f(0) = b$, $f(x)\geq 0$
  2. $f(x)$ is a continuous and non-increasing function

Let $g(x) = xf(x) + k(ab - \int_0^a f(x) dx)$.

Does there exist a $m$, such that $\max(g(x)) \geq mab$ for every $f$? If so, how does $m$ related to $k$? I'm assuming $m = h(k)$ for some function $h$ independent of $a$ and $b$, but I might be wrong.

Graphically this means, given a function $f(x)$

enter image description here

The area of $x f(x)$ and the area between $h(x)=b$ and $f(x)$ times $k$ is always larger than the area of the rectangle times $m$.

I don't know what area of mathematics deals with this kind of problem. Please retag if this is not analysis related.

$k=0$ then $m=0$ is the only thing I know. I have problem even proving the case when $k = 1$. I conjecture it's $m=3/4$ when $k=1$.

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1 Answer 1

up vote 3 down vote accepted

First, let's work out the possible pairs $\phi(f) = ( \sup_x \{xf(x)\}, \int_a^b f(t)dt)$ when $f$ is such a function.

Suppose $xf(x)$ takes some value $z \leq ab$. Then the smallest possible value for the integral of $f$ is $z$ :

Pick $x \leq a$ and $y \leq b$ such that $z = xy$. For any $\varepsilon > 0$, we can build a piecewise affine function $f$ by choosing $f(0) = b, f(\varepsilon) = f(x) = y, f(x+\varepsilon) = f(b) = 0$. Then, $\phi(f) = (z,z+b\varepsilon /2)$. Thus we can build $f$ with $\phi(f)$ as close to $(z,z)$ as we want.

Conversely, if $z = \sup_x xf(x)$, the highest possible value for the integral of $f$ is for $f(x) = \inf (b, z/x)$. We can compute the integral, and find $z(1 - \ln(z/ab))$.

Therefore, we know that the image of $\phi$ is $\{(z,t), 0 \leq z \leq t \leq z(1 - \ln(z/ab)), z \leq ab\}$. $$ \begin{array}{rcl} m & = & \frac{1}{ab} \inf_f (\sup_x g(x))) = k + \frac{1}{ab} \inf_f (\sup_x (xf(x)) - k\int_0^a f(t) dt) \\ &=& k + \frac{1}{ab} \inf_{(z,t) \in \text{Im}\phi} (z-kt)\end{array}$$

If $k\leq 0$, we get $m = k + \frac{1}{ab} \inf_z z(1-k) = k$

If $k\geq 0$, we get $m = k + \inf_z \frac{z}{ab} (1-k(1- \ln(z/ab))) = k + \inf_{u \in [0,1]} u(1-k(1- \ln(u)))$. The minimum occurs for $z/ab = u = e^{-1/k}$, which leads to $m = k(1-e^{-1/k})$.

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