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Suppose $ f $ is a continuous function from $ [0, 1] $ to $ [-1, 1] $ with $ |f(x)|\leq x, x\in [0, 1] $. Find the maximal value for $$ \left| \int_{0}^{1}(f(x))^2-f(x)dx \right| .$$

I have tried the following:

\begin{align*} \left| \int_{0}^{1}(f(x))^2-f(x)dx \right|&=\left| \int_{0}^{1}(f(x)-\frac{1}{2})^2-\frac{1}{4})dx \right| \\ &=\left| \int_{0}^{1}(f(x)-\frac{1}{2})^2dx-\frac{1}{4}\right| \end{align*} So we have to find some $ f(x) $ to make $ \int_{0}^{1}(f(x)-\frac{1}{2})^2dx $ as far from $ \frac{1}{4} $ as possible. Then I am stuck...

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    $\begingroup$ Just look the integrand and you can deduce the answer: when is $\int f(x)^2$ maximum, when is $\int -f(x)$ maximum? $\endgroup$
    – Winther
    Sep 23, 2018 at 9:06
  • $\begingroup$ @Winther How am I supposed to consider $ \int f(x)^2 $ and $\int -f(x) $ at the same time? $\endgroup$
    – Bach
    Sep 23, 2018 at 9:20
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    $\begingroup$ Consider them individually. For example $\int f(x)^2{\rm d}x$ is maximized by making $f^2$ as large as possible. With the given constraint this means $f(x) = x$ or $f(x) = -x$. You can also proceed from your derivation: you need to make $(f(x)- 1/2)^2$ either as large or as small as possible. With the given constraints this gives you just two functions to check. $\endgroup$
    – Winther
    Sep 23, 2018 at 9:21
  • $\begingroup$ Hint: Can you choose values for $f(x)$ so there is no cancellation? $\endgroup$ Sep 23, 2018 at 9:22

2 Answers 2

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Let $A=\{f\in \mathcal C([0,1],[-1,1])\;,\; \forall x, |f(x)|\leq x\}$. It is not hard to prove that $$\sup_{f\in A}\left|\int_0^1f^2-f \right|= \max\left(\sup_{f\in A}\left[\int_0^1f^2-f\right], -\inf_{f\in A}\left[\int_0^1f^2-f\right] \right)$$

Besides, for $f\in A$, $f^2(x)\leq x^2$, thus $\int_0^1 f^2-f\leq \int _0^1(x^2+x) dx=\frac 56$, and this upper bound is attained for $f(x)=-x$, hence $\sup_{f\in A}\left[\int_0^1f^2-f\right] = \max_{f\in A}\left[\int_0^1f^2-f\right]=\frac 56$.

Note also that $\int f^2-f\geq -\int_0^1 f\geq -\int_0^1 x=-\frac 12$. Hence $\inf_{f\in A}\left[\int_0^1f^2-f\right]\geq -\frac 12$ and $$\max\left(\sup_{f\in A}\left[\int_0^1f^2-f\right], -\inf_{f\in A}\left[\int_0^1f^2-f\right] \right) = \sup_{f\in A}\left[\int_0^1f^2-f\right] = \max_{f\in A}\left[\int_0^1f^2-f\right]=\frac 56$$ thus $$\sup_{f\in A}\left|\int_0^1f^2-f \right|=\frac 56$$ and the upper bound is attained for $x\mapsto -x$.

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    $\begingroup$ Good job! This one is relatively more rigorous. $\endgroup$
    – Bach
    Sep 23, 2018 at 10:29
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Consider the function $g(x) = x^2 - x$ defined on the interval $[-a, a]$, where $0 \le a \le 1$. Where is the maximum of $g(x)$? What is this maximum value?

Since $g(x)$ is a convex parabola, with only local minima in the interior of the interval the maximum must occur at an endpoint, i.e. at $a$ or $-a$. Comparing, we have, $$g(-a) = a^2 + a \ge a^2 - a = g(a),$$ so the maximum value occurs at $x = -a$.

Now, if $f : [0, 1] \to [-1, 1]$, with $|f(x)| \le x$ for all $x \in [0, 1]$, then $$f(x)^2 - f(x) \le x^2 + x,$$ using the above with $a = x$. This maximum is achieved when $f(x) = -x$. Given this pointwise inequality, it follows that $$\int_0^1 f(x)^2 - f(x) \; \mathrm{d}x \le \int_0^1 x^2 + x \; \mathrm{d}x = \frac{5}{6}.$$

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