Spring 2018, problem 68

Let $f, g\colon [0,1] \rightarrow (0,+\infty)$ be non-equal, continuous functions such that $\int_{0}^{1}f(x)dx = \int_{0}^{1}g(x)dx$. Let $y_n=\int_{0}^{1}{\frac{f^{n+1}(x)}{g^{n}(x)}dx}$, for $n\geq \mathbf{N}$. Prove that $(y_n)$ is an increasing and divergent sequence.


3 years ago

Let $f(x), g(x)$ be two nonequal, positive-valued, & continuous functions over $x \in [0,1]$ with equal areas:

$\int_{0}^{1} f(x) dx = \int_{0}^{1} g(x) dx$ (i)

If $f(x) > g(x)$ for all $0 \le x \le 1$, then:

1) $f(x) \cap g(x) = \varnothing$

2) $\int_{0}^{1} f(x) dx > \int_{0}^{1} g(x) dx$

and we have a contradiction to (i). A contradiction also exists for: $f(x)$ < $g(x)$ $\Rightarrow \int_{0}^{1} f(x) dx$ < $\int_{0}^{1} g(x) dx $. Therefore, $f(x)$ and $g(x)$ must intersect at a minimum of 1 point and up to as many as $N$ points (where $N \in \mathbb{N}$) within $[0,1]$ in order for (i) to be valid.

Let $0 \le x_1$ < $x_2$ < ... < $x_N \le 1$ be the intersection points of $f(x)$ and $g(x)$ such that:

$f(x) \le g(x)$ for $[0, x_1]; [x_2, x_3]; .... [x_{N-1}, x_N]$ (ii)

$f(x) > g(x)$ for $(x_1, x_2); (x_3, x_4); .... (x_N, 1]$ (iii)

and let us define the sequence:

$y_n = \int_{0}^{1} \frac{f^{n+1}(x)}{g^{n}(x)} dx = \int_{0}^{x_1} (\frac{f(x)}{g(x)})^n \cdot f(x) dx + \int_{x_1}^{x_2} (\frac{f(x)}{g(x)})^n \cdot f(x) dx + ... +\int_{x_{N-1}}^{x_N} (\frac{f(x)}{g(x)})^n \cdot f(x) dx + \int_{x_N}^{1} (\frac{f(x)}{g(x)})^n \cdot f(x) dx$ (iv)

As $n \rightarrow \infty$, we obtain at least one unbounded integral value in (iv) since $|\frac{f(x)}{g(x)}| > 1$ over any of the intervals belonging to (iii) $\Rightarrow (\frac{f(x)}{g(x)})^n \rightarrow \infty$.

Hence, ${y_n}$ is a divergent sequence.

$\mathbb{Q}. \mathbb{E}. \mathbb{D}.$

Again not so clear. How do you know f and g interesect at finitely many points. It is not clear why you need extra power of 'f' as well which is very important. Also divergent sequence does not mean monotonically increasing.

chorgeshashank007 3 years ago
3 years ago

This is similar to TME's solution, but, in addition, proves that the series $(y_n)$ is increasing. I think TME's proof shows that for $y_n = \int_{0}^{1} ( \frac{f^{n}(x)} {g^{n}(x)} ) dx$ the series $(y_n)$ is divergent. The extra factor $f(x)$ in the numerator is not needed for that proof. Hence for $y_n = \int_{0}^{1} ( \frac{g^{n}(x)} {f^{n}(x)} ) dx$ the series ($y_n)$ is also divergent. I believe the extra factor $f(x)$ in the numerator is needed to show that the series in the problem is increasing.

Start with $$\int_{0}^{1} (f(x)-g(x)) dx = 0, (i)$$ and then break up the integral into $\int_{+}dx$ which covers the regions where $f(x) - g(x) > 0$ and $\int_{-}dx$ which covers the regions where $f(x) - g(x) \lt 0$.

Then $$\int_{+}(f(x)-g(x)) dx = \int_{-}(g(x)-f(x)) dx. (ii)$$

The consecutive elements of the series $y_n$ differ by $$y_{n}-y_{n-1}
= \int_{0}^{1} ( \frac{f^{n}(x)}{g^{n}(x)})(f(x)-g(x))dx$$ or $$ y_{n}-y_{n-1}=\int_{+} ( \frac{f^{n}(x)}{g^{n}(x)})(f(x)-g(x))dx-\int_{-} ( \frac{f^{n}(x)}{g^{n}(x)})(g(x)-f(x))dx.(iii) $$

Use the fact $(\frac{f^{n}(x)}{g^{n}(x)})>1$ in $\int_{+}$ and $(\frac{f^{n}(x)}{g^{n}(x)}) \lt 1 $ in $\int_{-}$. Then starting from the equality in $(ii)$, we see the the first term ($\int_{+}$) in $(iii)$ is larger than the absolute value of the second term ($\int_{-}$) for all $n>0$. Thus $y_{n}-y_{n-1}$ is always positive which demonstrates that the series is always increasing and therefore also divergent.

3 years ago

How can you assume g(x)? g(x) can be anything.

f and g are independent . How can you say f(x)= g(1-x)?? I will post a correct solution soon. But your solution is wrong

chorgeshashank007 3 years ago