Let $f$ be a $C^2$ function on $[0,1]$ such that

$f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that


Find all $f$ for equality to occur.


First a variational argument. Assume that this expression is minimal for some smooth function $f$. Then let $\delta: [0,1] \to \mathbb{R}$ be twice differentiable and such that $\delta(0) = \delta'(0) = \delta'(1) = \delta(1) = 0$ and consider $f + t \cdot \delta$ for a real number $t$. This function also satisfies the boundary conditions, so

$$ \int_0^1 \left(f''(x) + t\cdot \delta''(x)\right)^2 dx $$

is minimal for $t = 0$. This implies that

$$ \int_0^1 f''(x) \delta''(x) dx = 0. $$

Apply partial integration twice to get

$$ \int_0^1 f^{(4)}(x) \delta(x) dx = 0. $$

Since this must hold for any such function $\delta$ it follows that $f^{(4)}$ is identically zero on $[0,1]$ and so must be a polynomial of degree at most three. The only such polynomial that satisfies the boundary conditions is

$$ f(x) = x^2 (x - 1). $$

For this $f$ you obtain the lower bound

$$ \int_0^1 \left(f''(x)\right)^2 dx = 4. $$

This argument has produced a nice candidate minimal function. Once this candidate is found the claim follows easily. Let $f$ be the polynomial above and take $g \in C^2[0,1]$ satisfying the boundary conditions as stated in the problem. Then $g = f + (g - f)$ and if we define $\delta = g - f$ then $\delta$ has the properties assumed above. In particular, by partial integration, we know that

$$ \int_0^1f''(x)\delta''(x) dx = 0. $$


$$ \int_0^1\left(g''(x)\right)^2 dx = \int_0^1\left(f''(x) + \delta''(x)\right)^2 dx = 4 + \int_0^1 \left(\delta''(x)\right)^2 dx \geq 4 $$

with equality only for $\delta'' = 0$. The latter implies $\delta = 0$ since $\delta'(0) = \delta(0) = 0$. So equality only holds when $g = f$.

  • $\begingroup$ Does it matter that we only know $f$ to be $C^2$? $\endgroup$ – mjqxxxx Oct 25 '12 at 21:43
  • $\begingroup$ @mjqxxxx Sure it matters. It means my argument doesn't apply. And I have to edit it a bit anyway. Thanks for pointing this out. $\endgroup$ – WimC Oct 26 '12 at 4:44
  • $\begingroup$ Fixed it... I hope. :-) $\endgroup$ – WimC Oct 26 '12 at 18:34

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