Find the maximum value of $\int_0^1 (f(x))^3 dx$, under certain conditions on $f(x)$ and $\int_0^1f(x)dx$

Question

Problem

I have gone through the solution here Finding the maximum value of $\int_0^1 (f(x))^3 dx$, given certain conditions on $f(x)$ and $\int_0^1 f(x) dx$ and satisfied with it but I am unable to find out the mistake of this solution available on the internet.Solution

Why is it wrong?

(Forgive me for using images. I am new here and not quite familiar with MathJax.)

Answer 1

It is correct to write that $$\int_0^1(f(x))^3dx\leq 1$$ but that does not mean that the maximum value is 1. The point is that the maximum value must be reached by the function at some value of x.

For eg. $(x-1)^2+6\geq 2$ is a correct statement for all x, but the minimum value of the function is not 2, as you can observe.

Again, $\sin x\leq 55$ is a correct statement, but the maximum value of $\sin x$ is 1, not 55.

In other words, the global maxima (if finite) must be a least upper bound of the range set of the function. In the previous example, 55 is just an upper bound.

P.S.

Now, you may wonder what makes $\frac 14$ so special? Just like we found a lower number than 1, can’t we find a lower number than $\frac 14$? NO. That’s because the function $$f(x)=\begin{cases} -\frac12, & \text{$x\in [0,\frac23)$}\\ 1, & \text{$x\in [\frac23,1]$}\end{cases}$$ satisfies all the criteria and has $\displaystyle \int_0^1(f(x))^3dx=\frac14$.