Spring 2018, problem 68

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.

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

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.

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