I have a function $f(x)$ which is defined and continuous on $[0, 1]$ and $f(0) = 0$. Also the derivative of the function is continuous on the $(0, 1)$. And I need to prove that if $\exists \ M$ such that the function $\sqrt{f^2(x) + f'^2(x)}$ is bounded i.e. $\sqrt{f^2(x) + f'^2(x)} \le M$ than $f(x) \le M$.
The fact that $f$ is bounded is obvious because it's continuous on the compact set, but the fact that that espression with a square root can be bounded by the same number as the function doesn't seem obvious. I've tried to use MVT but the only thing I came to is that $\forall \ x \in (0, 1]\ \ \exists \ \xi \in (0, x) : \frac{f(x)}{x} = f'(\xi)$ in particular $f(1) = f'(\xi)$ and I don't know what to do further (though I'm almost sure that I'm thinking the wrong way).
Would be very grateful for any hints! Thank in advance.
I think what you look for is simply the inequality $$ |a|\leq\sqrt{a^2+b^2}, $$ which is an obvious consequence of $a^2\leq a^2+b^2$.