Prove the integral is in the range of f

[Vardaan Bhat]

Homework Statement

If $f: [0,1] \rightarrow \mathbb{R}$ is continuous, show that $(n+1) \int_0^1 x^n f(x) \mathrm{d}x$ is in the range of $f$

Homework Equations

$(n+1) \int_0^1 x^n f(x) \mathrm{d}x=\int_0^1 (x^{n+1})' f(x) \mathrm{d}x$

The Attempt at a Solution

I tried integration by parts, but that led me into an endless loop. I have no other ideas/strategies to solve this.

Hints would be appreciated.

 

[Samy_A]

Homework Statement

If $f: [0,1] \rightarrow \mathbb{R}$ is continuous, show that $(n+1) \int_0^1 x^n f(x) \mathrm{d}x$ is in the range of $f$

Homework Equations

$(n+1) \int_0^1 x^n f(x) \mathrm{d}x=\int_0^1 (x^{n+1})' f(x) \mathrm{d}x$

The Attempt at a Solution

I tried integration by parts, but that led me into an endless loop. I have no other ideas/strategies to solve this.

Hints would be appreciated.

Integration by parts won't work, as you don't know if f is differentiable.

Hint 1:
As f is continuous on the closed interval [0,1], it has a minimum m and a maximum M.
That means, for example, that $(n+1) \int_0^1 x^n f(x) \mathrm{d}x \leq M (n+1) \int_0^1 x^n \mathrm{d}x$.
Hint 2: intermediate value theorem.

 

[Vardaan Bhat]

Why is the inequality in the first hint true? Shouldn't it just be M without the coefficient?

Also, for hint 2, I've tried this with just $f(m)\le f(x) \le f(M)$, integrating all the sides to yield $f(m)\le \int_0^1 f(x) \mathbb{d}x \le f(M)$. I don't know how to incorporate the $x^n$ though...

 

[Samy_A]

Why is the inequality in the first hint true? Shouldn't it just be M without the coefficient?

$f(x) \leq M$ for all $x \in [0,1]$, so that $x^nf(x) \leq x^n M$.
There is no coefficient in $M (n+1) \int_0^1 x^n \mathrm{d}x$, it is just the number $M$ multiplied by the number $n+1$.
Let's rewrite it as $(n+1) \int_0^1 x^n f(x) \mathrm{d}x \leq (n+1) M\int_0^1 x^n \mathrm{d}x$ for clarity.

Also, for hint 2, I've tried this with just $f(m)\le f(x) \le f(M)$, integrating all the sides to yield $f(m)\le \int_0^1 f(x) \mathbb{d}x \le f(M)$. I don't know how to incorporate the $x^n$ though...

The correct inequality is $m \leq f(x) \leq M$, for all $x \in [0,1]$.
You could multiply all terms in this inequality by a convenient function before integrating.
Or solve the integral you get in hint 1, both methods are equivalent.

 

[Vardaan Bhat]

So for all $x \in [0,1]$, we know that $m \le f(x) \le M$. But if we multiply each side by $(x^{n+1})'$ and integrate, that doesn't really help, does it? If we're multiplying by the function $(x^{n+1})'$ , does that mean we're multiplying by the value of the function for the $k$ at that point? In other words, when you say, multiply each side of the inequality $f(x_m) \le f(x) \le f(x_M)$ by the function $(x^{n+1})'$, do you mean we end up with $x_m^nf(x_m) \le x^nf(x) \le x_M^nf(x_M)$? And then we integrate over the interval $[0,1]$? But where does that leave us? It doesn't seem to help much...

 

[Samy_A]

So for all $x \in [0,1]$, we know that $m \le f(x) \le M$. But if we multiply each side by $(x^{n+1})'$ and integrate, that doesn't really help, does it? If we're multiplying by the function $(x^{n+1})'$ , does that mean we're multiplying by the value of the function for the $k$ at that point? In other words, when you say, multiply each side of the inequality $f(x_m) \le f(x) \le f(x_M)$ by the function $(x^{n+1})'$, do you mean we end up with $x_m^nf(x_m) \le x^nf(x) \le x_M^nf(x_M)$? And then we integrate over the interval $[0,1]$? But where does that leave us? It doesn't seem to help much...

Why don't you try, and see where it gets you?

You have $\forall x \in [0,1]: m \leq f(x) \leq M$.
Multiplying by $x^n$ (as you indeed suggested) gives: $\forall x \in [0,1]: mx^n \leq x^nf(x) \leq Mx^n$.

What do you get when you integrate over the interval [0,1]?
After integrating, remember the intermediate value theorem: any value between m and M lies in the range of f.

 

[Vardaan Bhat]

I'm not completely sure what you mean by multiply by $x^n$. Are you talking about multiplying by $x_m^n$, where $x_m$ is the value such that $f(x_m)=m$, or is this the same $x$ in the middle part of the inequality?

 

[Vardaan Bhat]

sorry, disregard the above post.

 

[Vardaan Bhat]

I followed your advice, ending up with $$mx^{n+1} \le (n+1)\int_0^1 x^nf(x) \mathbb{d}x \le Mx^{n+1}$$. Since $0 \le x \le 1$, I see how this makes the middle term less than (or equal to) the maximum, but it doesn't help us show that it's larger than (or equal to) than the minimum

 

[Samy_A]

I followed your advice, ending up with $$mx^{n+1} \le (n+1)\int_0^1 x^nf(x) \mathbb{d}x \le Mx^{n+1}$$. Since $0 \le x \le 1$, I see how this makes the middle term less than (or equal to) the maximum, but it doesn't help us show that it's larger than (or equal to) than the minimum

No, you did not correctly follow up my advice.
$M(n+1)\int_0^1 x^n dx \neq Mx^{n+1}$
$M(n+1)\int_0^1 x^n dx$ can't depend on some variable $x$.
$\int_0^1 x^n dx$ is a definite integral and thus a plain real number.

Solve it correctly and you will get an inequality $(n+1)\int_0^1 x^nf(x) \mathbb{d}x \leq \text{some number}$.

Doing the similar operation for $m$ will yield: $\text{some other number} \leq (n+1)\int_0^1 x^nf(x) \mathbb{d}x$.

 

[Vardaan Bhat]

I ended up with:

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Since we know that $f$ is continuous, we recognize that for all $x \in [0,1]$, $f(x)$ lies between the maximum and minimum values of $f$ over that domain. In other words, $$m \le f(x) \le M$$, where $m$ is the minimum of $f$ over $[0,1]$, and $M$ is the maximum over that same interval.

As a result, if we wish to prove that $(n+1) \int_0^1 x^n f(x) \mathrm{d}x$ is in the range of $f$, we must prove that $$m \le (n+1) \int_0^1 x^n f(x) \mathrm{d}x \le M.$$

Multiplying each side by $x^n$, we have $$x^nm \le x^nf(x) \le x^nM$$ for all $x \in [0,1]$.

Applying the Racetrack Theorem in "reverse," we can integrate each side of the inequality over the interval $[0,1]$, yielding $$\int_0^1 x^nm \le \int_0^1 x^nf(x) \le \int_0^1 x^nM.$$ Using basic integration techniques, we can rewrite this as $$\frac{m}{n+1} \le \int_0^1 x^nf(x) \le \frac{M}{n+1}.$$ Multiplying each side by $n+1$, we now have $$m \le (n+1) \int_0^1 x^n f(x) \mathrm{d}x \le M.$$

Since $$(n+1) \int_0^1 x^n f(x) \mathrm{d}x$$ lies between the minimum and maximum values of $f$, it must be in the range of $f$, and therefore our proof is complete.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 

[Vardaan Bhat]

Thanks for the help!