Question. $f$ is real valued function on $[0,1]$ such that $(2x-1)(2f(x)-x)>0$ for all $x≠1/2$ then prove that if $f'(1/2)$ exist then it cannot be smaller than $1/2$.
Here is my argument.
Let us assume that $f'(1/2)$ exists. Then, by the definition of the derivative, we have
$f'(1/2)=\lim\limits_{x\to1/2}\dfrac{f(x)-f(1/2)}{x-1/2}$
We can rewrite the given inequality as
$2f(x) - x > 0$ if $x < 1/2$
$2f(x) - x < 0$ if $x > 1/2$
Since $(2x - 1) > 0$ for $x > 1/2$ and $(2x - 1) < 0$ for $x < 1/2$, we can infer that
$2f(x) - x > 0$ for $x > 1/2$
$2f(x) - x < 0$ for $x < 1/2$
Now, let us consider a sequence $(x_n)$ that converges to $1/2$ from below, i.e., $x_n < 1/2$ for all $n$ and $\lim\limits_{n\to\infty}x_n=1/2$.
Then, we have
$2f(x_n) - x_n > 0$
Multiplying both sides by $2x_n - 1$, we get
$(2x_n - 1)(2f(x_n) - x_n) > 0$
Dividing both sides by $x_n - 1/2$, we get
$2f(x_n) - x_n > 0$ for $n$ sufficiently large
Therefore, for $x$ close to $1/2$, $f(x)$ is greater than $x/2$. This implies that the right-hand limit of $f(x)/x$ as $x$ approaches $1/2$ is at least $1/2$. In other words,
$\lim\limits_{x\to(1/2)^+}\dfrac{f(x)}x\geqslant\dfrac12$
By the definition of the derivative, we have
$f'(1/2)=\lim\limits_{x\to1/2}\dfrac{f(x)-f(1/2)}{x-1/2}=\lim\limits_{x\to1/2}\dfrac{f(x)}x$
Since the left-hand limit of $f(x)/x$ as $x$ approaches $1/2$ is also at least $1/2$ (because $f(x)$ is non-negative for $x$ close to $1/2$), we have
$\lim\limits_{x\to(1/2)^-}\dfrac{f(x)}x\geqslant\dfrac12$
Therefore, $f'(1/2)$ is at least $1/2$, as required to be shown. Please cheak this or give another solution... thank you...
Since for hypothesis there exists $\,f’\left(\dfrac12\right)\,,\,$ the function $\,f(x)\,$ is continuous at $\,x=\dfrac12\,.$
Moreover,
$ f(x)>\dfrac x2\quad$ for $\;x>\dfrac12\;,$
$ f(x)<\dfrac x2\quad$ for $\;x<\dfrac12\,.$
Consequently,
$f\left(\dfrac12\right)=\lim\limits_{x\to(1/2)^+}f(x)\geqslant\lim\limits_{x\to(1/2)^+}\dfrac x2=\dfrac14\;,$
$f\left(\dfrac12\right)=\lim\limits_{x\to(1/2)^-}f(x)\leqslant\lim\limits_{x\to(1/2)^-}\dfrac x2=\dfrac14\;,$
hence $\;f\left(\dfrac12\right)=\dfrac14\;.$
Furthermore ,
$\dfrac{f(x)-f\left(\frac12\right)}{x-\frac12}=\dfrac{f(x)-\frac14}{x-\frac12}=$
$=\dfrac{f(x)-\frac x2}{x-\frac12}+\dfrac{\frac x2-\frac14}{x-\frac12}=\dfrac{2f(x)-x}{2x-1}+\dfrac12=$
$=\dfrac{(2x-1)[2f(x)-x]}{(2x-1)^2}+\dfrac12>\dfrac12\quad$ for all $\;x\ne\dfrac12\;.$
Consequently,
$f’\left(\dfrac12\right)=\lim\limits_{x\to\frac12} \dfrac{f(x)-f\left(\frac12\right)}{x-\frac12}\geqslant\dfrac12\;.$