Counting number of $f(x)$

Question

Let $S$ be the set of continuous functions $f:[0,1]\to[0,\infty)$ that satisfy $$\int_{0}^{1} x^2f(x) dx=\frac{1}{2}\int_{0}^{1} xf^2(x) dx+\frac{1}{8}.$$ Is there at most one element in the set $S$ ?

I have seen that $f(x)=x$ satisfies the above equation and no other $f(x)$ of the form $f(x)=ax$ or $f(x)=ax^2$ satisfies this .

Answer 1

I'm sure I am missing some rigor here, but here is an attempt that shows that no such other function exists. Suppose it does and consider $f(x) = x + g(x)$. All such continuous functions could be written this way (suppose $g(x) = h(x) - x$ to see this). Inserting this into the expression and simplifying, we have $$ \int_0^1 x [g(x)]^2 \ dx = 0. $$ Since $g(x)$ is a continuous function, it must be that $g(x) \equiv 0$.

Answer 2

Yes.

If $f$ satisfies the equality, then \begin{align*} \int_0^1 (f(x)-x)^2x dx &= \int_0^1 f^2(x)x + x^3 -2x^2f(x) dx \\ & = \int_0^1 x^3 dx +2 \left[\dfrac{1}{2}\int_0^1 f^2(x)x dx-\int_0^1 x^2f(x) dx \right] \\ & = \dfrac{1}{4}-2\times \dfrac{1}{8} \\ & = 0 \end{align*}

Since the integrand is positive and continuous, one deduces that for every $x \in [0,1]$, $(f(x)-x)^2x=0$, i.e. that $f(x)=x$ for every $x \in (0,1]$, and since $f$ is continuous, $$\boxed{\forall x \in [0,1],\ f(x)=x}$$

Answer 3

We have given

$\displaystyle \int^1_0 x^2f(x)dx=\frac{1}{2}\int^1_0x(f(x))^2dx+\frac{1}{8}$

We have given

$\displaystyle \int^1_0\bigg(x^2f(x)-\frac{x(f(x))^2}{2}\bigg)dx=\frac{1}{8}$

Now using

$\displaystyle \int^1_0 2\bigg[x\frac{f(x)}{2}\bigg(x-\frac{f(x)}{2}\bigg)\bigg]dx\leq \frac{1}{4}\int^1_02x\bigg[\frac{f(x)}{2}+x-\frac{f(x)}{2}\bigg]^2dx=\frac{1}{8}$

Equality hold when

$\displaystyle \frac{f(x)}{2}=x-\frac{f(x)}{2}\Longrightarrow f(x)=x$

(Above we have used $\displaystyle ab\leq \frac{1}{4}(a+b)^2$

And equality hold when $a=b$)