Comparing $\|f\|$ with $\|f''\|$ on an open interval So I was looking at one of my dad's old math exams and came across the following question:
Suppose $f\in C^2((a,b))$ (open interval) and for some $a <x_1 <x_2 <b$ we have $f(x_1) =0 =f(x_2)$. Prove $\|f\| \le (b-a)^2 \|f''\|$ where $\|g\| = \sup_{x \in (a,b)} |g(x)|$.
My first inclination was a FTC argument, but of course this can't work (for example, $1/x$ on $(0,1)$). I also considered the following:
By Rolle's, there exists $c \in (x_1, x_2)$ such that $f'(c) =0$. Then Taylor expanding, we have $f(x) = f(c) + \displaystyle \frac12 (x-c)^2 f''(c) +g(x-c)$, where $g(x-c) \to 0$ as $x\to c$, but this was also fruitless. 
Any ideas? I feel like the problem must use Rolle's theorem somehow, but I'm not sure how. 
 A: Mean value theorem:
For any $x,y\in (a,b)$ there exists a $c \in (x,y)$ s.t.
$(x-y)f'(c) = f(x) - f(y)$
which implies 
$|f(x) - f(y)|\le |x-y|\|f'(x)\|  $
and since $|x-y| < b-a$
$|f(x) - f(y)| \le \|f'(x)\| (b-a)$ for all $x,y \in (a,b)$ 
$\|f(x)\| - f(x_1) = \|f(x)\| \le \|f'(x)\| (b-a)$ 
similarly...
$\forall x,y \in (a,b): |f'(x) - f'(y)| \le \|f''(x)\|(b-a) $
By Rolle's theorem we know that there exists some $x^*$ such that $f'(x^*) = 0$
$\|f'(x)\| \le \|f''(x)\|(b-a)\\
\|f'(x)\|(b-a) \le \|f''(x)\|(b-a)^2\\
\|f(x)\|\le \|f'(x)\|(b-a) \le \|f''(x)\|(b-a)^2$
A: One minor point, $\| g \| = \sup_{x \in (a,b)} \lvert g(x) \rvert$ isn't actually a norm on $C^2((a,b))$ since it won't give a finite value for every continuous function. Consider, $g(x) = 1/x$ for $x \in (0,1)$ is a infinitely differentiable function where each derivative is unbounded. This is why we usually only deal with $C(\Omega)$ for compact $\Omega$. 
Regardless, if we consider $[a,b]$, Doug M posted a perfectly good solution using the mean value theorem. I just wanted to note that you can also do this using the fundamental theorem of calculus. Note that since $f(x_1) = 0$ for any $x \in [a,b]$, we have $$f(x) = \int_{x_{1}}^x f'(t) dt$$ whence $$\lvert f(x) \rvert \le \int^x_{x_1} \lvert f'(t) \rvert dt$$ [here some care is needed. For example, if $x  < x_1$, then when we pass the absolute values inside the integral, we would need to switch the bounds so that the result is positive. I will leave it as is for the sake of demonstration]. Next, note that since $f(x_2) = 0$, for all $t$, we have $$f'(t) = \int_{x_2}^{t} f''(s) ds$$ and so $$\lvert f'(t) \rvert \le \int^t_{x_2} \lvert f''(s) \rvert ds.$$ Compounding these bounds gives $$\lvert f(x) \rvert \le \int^x_{x_1} \int^t_{x_2} \lvert f''(s) \rvert ds dt.$$ But $\lvert f''(s) \rvert \le \| f'' \|$ for all $s$ so $$\lvert f(x) \rvert \le \int^x_{x_1} \int^t_{x_2} \|f'' \| dx dt = \| f'' \| \int^x_{x_1} (t-x_2) dt.$$ Now $(t- x_2) \le (b-a)$ so $$\lvert f(x) \rvert \le (b-a) \| f'' \| \int^x_{x_1} dt = (b-a)(x - x_1) \|f''\| \le (b-a)^2 \| f'' \|.$$ Since this bound holds pointwise, we can pass to the supremum and find $$\| f\| \le (b-a)^2 \| f'' \|$$ as desired. 
