Integrals problems and inequality

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tkd23112006

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tkd23112006

#1 h Feb 16, 2025, 1:32 PM • 3 Y

Y by PolyaPal, removablesingularity, Hairy_Ball_Theorem

Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$

This post has been edited 1 time. Last edited by tkd23112006, Feb 16, 2025, 1:33 PM
Reason: Incorrect format

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removablesingularity

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removablesingularity

#2 h Mar 14, 2025, 3:55 AM

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what a beautiful problem

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PolyaPal

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PolyaPal

#3 h Mar 15, 2025, 6:02 PM

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Solution?

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removablesingularity

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removablesingularity

#4 h Mar 18, 2025, 9:28 AM

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This is a problem using same assumption as in the given problem, but with different conclussion.
https://math.stackexchange.com/questions/221288/if-int-x1ftdt-ge-frac1-x22-x-in0-1-prove-that-int-01ft2dt-g

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PolyaPal

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PolyaPal

#5 h Mar 18, 2025, 3:26 PM

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I know this problem, but I can't see how we can solve the AoPS problem using the same ideas. Perhaps some form of Holder's inequality?

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Hairy_Ball_Theorem

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Hairy_Ball_Theorem

#6 h Mar 19, 2025, 1:29 AM

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I am not sure of this solution but it seems to work and note that we can exclude x=1 as it doesn't effect the integral and allows us to use (1-x) to divide at both sides

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removablesingularity

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removablesingularity

#7 h Mar 19, 2025, 7:14 AM

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Could you explain why $\left(1-x\right)f\left(x\right) \ge \int_x^1 f\left(t\right)dt$ ?

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Hairy_Ball_Theorem

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Hairy_Ball_Theorem

#8 h Mar 19, 2025, 9:40 AM

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Well i did assume that the function, f(x) is non-increasing and non-negative(which is already given ) on [0,1] thus it simply follows from that
$f(t)\le f(x) \quad \forall t\in [x,1]$ and so $\int_{x}^{1}f(t)dt\le\int_{x}^{1}f(x)dt=f(x)\int_{x}^{1}dt=(1-x)f(x)$ $\implies (1-x)f(x)\ge\int_{x}^{1}f(t)dt$

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removablesingularity

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removablesingularity

#9 h Mar 19, 2025, 1:24 PM

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Hairy_Ball_Theorem wrote:

Well i did assume that the function, f(x) is non-increasing and non-negative(which is already given ) on [0,1] thus it simply follows from that
$f(t)\le f(x) \quad \forall t\in [x,1]$ and so $\int_{x}^{1}f(t)dt\le\int_{x}^{1}f(x)dt=f(x)\int_{x}^{1}dt=(1-x)f(x)$ $\implies (1-x)f(x)\ge\int_{x}^{1}f(t)dt$

Actually I cannot see any non increasing assumption on the problem statement, could you elaborate?

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PolyaPal

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PolyaPal

#10 h Mar 19, 2025, 3:34 PM

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You can't assume that $f$ is either increasing or decreasing. However, it may be possible to show that $f(x) \geq x$ for all $x \in [0, 1]$ .

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Hairy_Ball_Theorem

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Hairy_Ball_Theorem

#11 h Mar 20, 2025, 8:53 AM

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removablesingularity wrote:

Hairy_Ball_Theorem wrote:

Well i did assume that the function, f(x) is non-increasing and non-negative(which is already given ) on [0,1] thus it simply follows from that
$f(t)\le f(x) \quad \forall t\in [x,1]$ and so $\int_{x}^{1}f(t)dt\le\int_{x}^{1}f(x)dt=f(x)\int_{x}^{1}dt=(1-x)f(x)$ $\implies (1-x)f(x)\ge\int_{x}^{1}f(t)dt$

Actually I cannot see any non increasing assumption on the problem statement, could you elaborate?

I mean yeah, as i said i assumed that part (of function being non-increasing) while the thing which was given is function is non-negative...

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PolyaPal

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PolyaPal

#12 h Mar 20, 2025, 8:08 PM

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You can't assume that. There must be a clever way to do this problem.

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MS_asdfgzxcvb

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MS_asdfgzxcvb

#13 h Mar 21, 2025, 12:55 PM • 2 Y

Y by removablesingularity, Alphaamss

tkd23112006 wrote:

Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$

$\emph{Proof.}$ All unlabelled integrals are in $[0,1]$ . We need to prove $\int f^n\geqslant \int x^{n-1}f,$ and we have $\int_x^1 f\geqslant \int_x^1 x,$ by IBP we have
$\begin{align*}\xRightarrow{\hspace{25pt}}\int x^{n-1}f&=0+(n-1)\int \left(x^{n-2}\int_x^1 f\right)\\ \xRightarrow{\hspace{25pt}}\int x^{n-1}f&\ge (n-1)\int x^{n-2}\cdot\left(\tfrac{1-x^2}2\right)=\tfrac 1{n+1}. \end{align*}$ Now by Holder,
$\begin{align*}\xRightarrow{\hspace{25pt}}&\left(\int x^n\right)^{n-1} \int f^n\ge \left(\int x^{n-1}f\right)^n \ge\left(\tfrac1{n+1}\right)^{n-1}\cdot\int x^{n-1}f\\ \color{blue}\xRightarrow{\hspace{25pt}}&\color{blue}\int f^n\ge \int x^{n-1}f.\qquad \color{black}\blacksquare \end{align*}$

This post has been edited 1 time. Last edited by MS_asdfgzxcvb, Mar 21, 2025, 12:57 PM

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PolyaPal

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PolyaPal

#14 h Mar 22, 2025, 7:45 PM

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A bit confusing without the variables of integration specified.

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removablesingularity

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removablesingularity

#15 h Mar 22, 2025, 11:04 PM

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PolyaPal wrote:

A bit confusing without the variables of integration specified.

But it works

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PolyaPal

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PolyaPal

#16 h Mar 22, 2025, 11:40 PM

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I don't see Holder's inequality used properly.

This post has been edited 1 time. Last edited by PolyaPal, Mar 22, 2025, 11:45 PM
Reason: I couldn't get the umlaut on the 'o' in the mathematician's name/

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Alphaamss

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Alphaamss

#17 h Mar 23, 2025, 2:46 AM • 2 Y

Y by teomihai, removablesingularity

MS_asdfgzxcvb wrote:

tkd23112006 wrote:

Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$

$\emph{Proof.}$ All unlabelled integrals are in $[0,1]$ . We need to prove $\int f^n\geqslant \int x^{n-1}f,$ and we have $\int_x^1 f\geqslant \int_x^1 x,$ by IBP we have
$\begin{align*}\xRightarrow{\hspace{25pt}}\int x^{n-1}f&=0+(n-1)\int \left(x^{n-2}\int_x^1 f\right)\\ \xRightarrow{\hspace{25pt}}\int x^{n-1}f&\ge (n-1)\int x^{n-2}\cdot\left(\tfrac{1-x^2}2\right)=\tfrac 1{n+1}. \end{align*}$ Now by Holder,
$\begin{align*}\xRightarrow{\hspace{25pt}}&\left(\int x^n\right)^{n-1} \int f^n\ge \left(\int x^{n-1}f\right)^n \ge\left(\tfrac1{n+1}\right)^{n-1}\cdot\int x^{n-1}f\\ \color{blue}\xRightarrow{\hspace{25pt}}&\color{blue}\int f^n\ge \int x^{n-1}f.\qquad \color{black}\blacksquare \end{align*}$

Nice job! Just give an explanation of the use of Hölder inequality:
$\left(\int_0^1 x^ndx\right)^{n-1} \int_0^1 f^n(x)dx\ge \left(\int_0^1 x^{n-1}f(x)dx\right)^n \iff \left(\int_0^1\left (x^{n-1}\right)^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}} \left(\int_0^1 f^n(x)dx\right)^{\frac1n}\ge \int_0^1 x^{n-1}f(x)dx.$

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