$|f(1)^2-f(0)^2|\le 2$ Let H the set of the functions $f:[0,1]\to\mathbb{R}$ differentiable, with $f'(x)$ continuous and $\int^1_0 (f(x)^2)\,dx =\int^1_0 (f'(x)^2)\,dx=1$
Prove that
$|f(1)^2-f(0)^2|\le 2$ for any $f\in H$.
I did not had any idea at it other than using the mean value theorem but I do not think it is helpful.
Because $(f^2)'=2ff'$, you have
$$f^2(1)-f^2(0) =  2\int_0^1 f(x)f'(x)dx $$
So
$$|f^2(1)-f^2(0)| =  2\left|\int_0^1 f(x)f'(x)dx \right| \leq 2 \left(\int_0^1 f^2(x) dx\right)^{1/2}\left(\int_0^1 f'^2(x) dx\right)^{1/2}$$
by Cauchy-Schwarz, so because
$$\int_0^1 f^2(x) dx=\int_0^1 f'^2(x) dx=1$$ you deduce
$$|f^2(1)-f^2(0)| \leq 2$$
You can't prove something that is not true. If you use in particular $f(x) = e ^x $ then
$$ \int_0^1 {f(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
$$ \int_0^1 {f'(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
So we can say:
$$ \int_0^1 {f(x)^2} dx =  \int_0^1 {f'(x)^2} dx $$
But:
$$ \vert{f(1)^2 - f(0)^2 }\vert = \vert e^2 - 1 \vert = 6.38 &gt; 2 $$