Given $a\in (0,1)$ and $C$ the set of increasing functions $f:[0,1]\to [0,\infty )$ such that $\int\limits_{0}^{1}{f(x)}dx=1$ . Determine: $(a)\underset{f\in C}{\mathop{\max }}\,\int\limits_{0}^{a}{f(x)dx}$ $(b)\underset{f\in C}{\mathop{\max }}\,\int\limits_{0}^{a}{{{f}^{2}}(x)dx}$
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Tags: function, integration, real analysis, real analysis unsolved
opptoinfinity
27.06.2018 10:32
Any solution ?
Catalin
27.06.2018 13:35
(a) Let $f$ be a function from $C.$ We will prove that $$\int_0^af(x) dx \leq a$$Indeed, $$a-\int_0^af(x) dx=a\int_0^1f(x) dx-\int_0^af(x) dx=a\int_a^1f(x) dx-(1-a)\int_0^af(x) dx \geq a\int_a^1f(a) dx-(1-a)\int_0^af(a) dx=0$$But for $f \equiv 1$ we have equality, thus the maximum is $a.$
(b) The maximum is $a,$ if $a \leq 1/2$ and $1/(4(1-a)),$ if $a>1/2.$ The first value is reached for $f \equiv 1$ and the second one for $$f(x)=\begin{cases} 0, & \text{if } 0 \leq x \leq 2a-1 \\ 1/(2(1-a)), & \text{if } 2a-1 < x \leq 1 \end{cases}$$Let $f$ be a function from $C.$ We then have $$\int_0^a (f(x))^2dx \leq f(a)\int_0^af(x) dx \leq \left(\frac{1}{1-a}\int_a^1f(x) dx \right) \left( \int_0^af(x) dx \right)=\frac{1}{1-a} \left( 1- \int_0^af(x) dx \right) \left(\int_0^af(x) dx \right)$$Note that $\max \{t(1-t) \mid t \leq a \} = \begin{cases}a(1-a), & \text{if } a \leq 1/2, \\ 1/4, & \text{if } a>1/2; \end{cases}$
In both cases, the maximum is attained only in one point: for $t=a,$ in the first case, and for $t=1/2,$ in the second one.
Hence, $$\int_0^a(f(x))^2 dx \leq \begin{cases}a, & \text{if } a \leq 1/2 \\ 1/(4(1-a)), & \text{if } a>1/2 \end{cases}$$
Raunii
10.03.2020 08:33
Catalin wrote:
(a) Let $f$ be a function from $C.$ We will prove that $$\int_0^af(x) dx \leq a$$Indeed, $$a-\int_0^af(x) dx=a\int_0^1f(x) dx-\int_0^af(x) dx=a\int_a^1f(x) dx-(1-a)\int_0^af(x) dx \geq a\int_a^1f(a) dx-(1-a)\int_0^af(a) dx=0$$But for $f \equiv 1$ we have equality, thus the maximum is $a.$
(b) The maximum is $a,$ if $a \leq 1/2$ and $1/(4(1-a)),$ if $a>1/2.$ The first value is reached for $f \equiv 1$ and the second one for $$f(x)=\begin{cases} 0, & \text{if } 0 \leq x \leq 2a-1 \\ 1/(2(1-a)), & \text{if } 2a-1 < x \leq 1 \end{cases}$$Let $f$ be a function from $C.$ We then have $$\int_0^a (f(x))^2dx \leq f(a)\int_0^af(x) dx \leq \left(\frac{1}{1-a}\int_a^1f(x) dx \right) \left( \int_0^af(x) dx \right)=\frac{1}{1-a} \left( 1- \int_0^af(x) dx \right) \left(\int_0^af(x) dx \right)$$Note that $\max \{t(1-t) \mid t \leq a \} = \begin{cases}a(1-a), & \text{if } a \leq 1/2, \\ 1/4, & \text{if } a>1/2; \end{cases}$
In both cases, the maximum is attained only in one point: for $t=a,$ in the first case, and for $t=1/2,$ in the second one.
Hence, $$\int_0^a(f(x))^2 dx \leq \begin{cases}a, & \text{if } a \leq 1/2 \\ 1/(4(1-a)), & \text{if } a>1/2 \end{cases}$$
Where did you find the official solution
Raunii
16.07.2021 15:43
Catalin wrote:
(a) Let $f$ be a function from $C.$ We will prove that $$\int_0^af(x) dx \leq a$$Indeed, $$a-\int_0^af(x) dx=a\int_0^1f(x) dx-\int_0^af(x) dx=a\int_a^1f(x) dx-(1-a)\int_0^af(x) dx \geq a\int_a^1f(a) dx-(1-a)\int_0^af(a) dx=0$$But for $f \equiv 1$ we have equality, thus the maximum is $a.$
(b) The maximum is $a,$ if $a \leq 1/2$ and $1/(4(1-a)),$ if $a>1/2.$ The first value is reached for $f \equiv 1$ and the second one for $$f(x)=\begin{cases} 0, & \text{if } 0 \leq x \leq 2a-1 \\ 1/(2(1-a)), & \text{if } 2a-1 < x \leq 1 \end{cases}$$Let $f$ be a function from $C.$ We then have $$\int_0^a (f(x))^2dx \leq f(a)\int_0^af(x) dx \leq \left(\frac{1}{1-a}\int_a^1f(x) dx \right) \left( \int_0^af(x) dx \right)=\frac{1}{1-a} \left( 1- \int_0^af(x) dx \right) \left(\int_0^af(x) dx \right)$$Note that $\max \{t(1-t) \mid t \leq a \} = \begin{cases}a(1-a), & \text{if } a \leq 1/2, \\ 1/4, & \text{if } a>1/2; \end{cases}$
In both cases, the maximum is attained only in one point: for $t=a,$ in the first case, and for $t=1/2,$ in the second one.
Hence, $$\int_0^a(f(x))^2 dx \leq \begin{cases}a, & \text{if } a \leq 1/2 \\ 1/(4(1-a)), & \text{if } a>1/2 \end{cases}$$
How do we know those values are indeed achieved in the second part, in the first part you gave the example when equality happens but what about second case
YII.I.
12.01.2024 22:48
A little but ineffective problem. Is the $C$ defined in the problem a set but not a proper class? (I never learned set theory. But I'm really curious about this)