Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,

$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$

holds.)

What values can the following integral take for these functions?

$$
\int_{0}^{1} f^{2}(x) d x
$$
a) One of the famous inequalities in integral calculus states that if the squares of the functions $f_{1}, f_{2}$ are also integrable, then

$$
\left(\int_{a}^{b} f_{1}(x) f_{2}(x) d x\right)^{2} \leqq \int_{a}^{b} f_{1}^{2}(x) d x \int_{a}^{b} f_{2}^{2}(x) d x
$$

(Cauchy-Schwarz inequality). Indeed, based on $f_{1}, f_{2}$, the function

$$
I(\lambda)=\int_{a}^{b}\left[f_{1}(x)+\lambda f_{2}(x)\right]^{2} d x=\int_{a}^{b} f_{1}^{2}(x) d x+2 \lambda \int_{a}^{b} f_{1}(x) f_{2}(x) d x+\lambda^{2} \int_{a}^{b} f_{2}^{2}(x) d x
$$

is non-negative for all $\lambda$. Since $I(\lambda)$ is a quadratic function of $\lambda$, this can only be true if the discriminant is non-positive, which is precisely the inequality (3). Applying (3) to the functions $f_{1}(x)=f(x), f_{2}(x)=1$, we get

$$
\int_{0}^{1} f^{2}(x) d x \geqq \int_{0}^{1} f(x) d x=1
$$

Thus, the smallest possible value of (2) is 1, which can be achieved (even under the constraint (1)) by choosing $f(x)=1$ (and only by this choice).

b) According to our assumption, $f(x)$ is concave from below, i.e., convex from above. Since $f(x) \geqq 0$, this means that the region $F$ bounded by the $x$-axis, the $y$-axis, and the graph of $f(x)$ is a convex set. Consider the horizontal chords of this set and associate with each chord a triangle whose base is the segment of the $x$-axis between 0 and 1, and whose sides pass through the endpoints of the chord. Denote the triangle corresponding to the chord of height $m$ by $H(m)$, the area of the part of $F$ not covered by $H(m)$ by $A(m)$, and the part of $H(m)$ not covered by $F$ by $B(m)$. Since $F$ is convex, if $m_{1}<m_{2}$, $H\left(m_{1}\right)$ contains $H\left(m_{2}\right)$, and $A\left(m_{1}\right) \leqq A\left(m_{2}\right)$, $B\left(m_{1}\right) \geqq B\left(m_{2}\right)$. The largest possible value of $m$ is equal to the maximum of $f(x)$, denoted by $M$. Clearly, $B(M)$ is 0, and since the area of $H(M)$ is less than the area of $F$, $M \leqq 2$. The smallest possible value of $m$ is the smaller of $f(0)$ and $f(1)$, and this value can be chosen as $m$. Here, the limit of $A(m)$ is 0, and as is easily seen, the functions $A$ and $B$ are continuous. Therefore, there exists an $m$ such that

$$
A(m)=B(m)
$$

and for this $m$, $H(m)=1$, meaning the third vertex of $H(m)$ lies on the line $y=2$. Denote the function formed by the two sides of $H(m)$ by $f_{0}$. For this function,

![](https://cdn.mathpix.com/cropped/2024_05_02_8629f7e329cb122734adg-1.jpg?height=727&width=444&top_left_y=1531&top_left_x=835)

$$
\begin{gathered}
\int_{0}^{1} f_{0}^{2}(x) d x-\int_{0}^{1} f^{2}(x) d x=\int_{0}^{1}\left[f_{o}(x)-f(x)\right]\left[f_{0}(x)+\right. \\
f(x)] d x \geqq 2 m \int_{0}^{1}\left[f_{0}(x)-f(x)\right] d x \geqq 0
\end{gathered}
$$

because

$$
\int_{0}^{1}\left[f_{0}(x)-f(x)\right]\left[f_{0}(x)+f(x)-2 m\right] d x \geqq 0
$$

since here the integrand is non-negative: where $f_{0} \leqq f$, both $f_{0}$ and $f$ are less than $m$, so the product of both factors is positive. Since $f_{0}$ is also concave from below, and its integral is 1, $f_{0}$ also belongs to the set of functions for which we are seeking the maximum value of (2). It is easily seen that the square integral of $f_{0}$ is $4 / 3$, so for the functions under consideration,

$$
1 \leqq \int_{0}^{1} f^{2}(x) d x \leqq \frac{4}{3}
$$