Bound on $f(0)^2$ by integrals of $f^2$ and $(f')^2$ on $[0,1]$.

Question

Let $f$ be a function which is $C^1((0,1))\cap C([0,1])$. I would like to be able to show $$ \frac{1}{2}f(0)^2 \leq \int_0^1 f(x)^2dx + \int_0^1f'(x)^2dx $$ where we are assuming that $f$ is a real-valued function.

We have attempted to use Young's inequality to reduce to $\frac{d}{dx}[f(x)^2]$, but this does not work.

Thanks in advance for any ideas!

edit: Counterexamples are welcome, obviously, but we believe this at the moment.

Answer 1

I think the inequality is correct. Let $$ f(x)=\sum_{n=0}^\infty(a_n\cos(n\pi x)+b_n\sin(n\pi x)). $$ This is because $\{\cos(n\pi x),\sin(n\pi x)\}_{n=0}^\infty$ is dense in $C^1((0,1))\cap C([0,1])$. Noting that, for integers $m, n$, \begin{eqnarray} &\int_0^1\cos^2(n\pi x)dx=\int_0^1\sin^2(n\pi x)dx=\frac12,\\ &\int_0^1\cos(n\pi x)\sin(m\pi x)dx=0,\\ &\int_0^1\cos(n\pi x)\cos(m\pi x)dx=0 \text{ if }m\neq n,\\ &\int_0^1\sin(n\pi x)\sin(m\pi x)dx=0 \text{ if }m\neq n, \end{eqnarray} we have \begin{eqnarray} \int_0^1(f(x))^2dx&=&\int_0^1(\sum_{n=0}^\infty(a_n\cos(n\pi x)+b_n\sin(n\pi x)))^2dx\\ &=&\int_0^1\sum_{m,n=0}^\infty(a_m\cos(m\pi x)+b_m\sin(m\pi x))(a_n\cos(n\pi x)+b_n\sin(n\pi x))dx\\ &=&\frac{1}{2}\sum_{n=0}^\infty (a_n^2+b_n^2),\\ \int_0^1(f'(x))^2dx&=&\int_0^1(\sum_{n=0}^\infty(-n\pi a_n\sin(n\pi x)+n\pi b_n\cos(n\pi x)))^2dx\\ &=&\int_0^1\sum_{m,n=0}^\infty mn\pi^2(-a_m\sin(m\pi x)+b_m\cos(m\pi x))(-a_n\sin(n\pi x)+b_n\cos(n\pi x))dx\\ &=&\frac{1}{2}\sum_{n=0}^\infty n^2\pi^2(a_n^2+b_n^2). \end{eqnarray} Thus \begin{eqnarray} &&\int_0^1(f(x))^2dx+\int_0^1(f'(x))^2dx-\frac12(f(0))^2\\ &=&\frac12\sum_{n=0}^\infty (n^2\pi^2+1)(a_n^2+b_n^2)-\frac12(\sum_{n=0}^\infty a_n)^2\\ &\ge&\frac12\left(\sum_{n=0}^\infty(n^2\pi^2+1)a_n^2-(\sum_{n=0}^\infty a_n)^2\right)+\frac12\sum_{n=0}^\infty(n^2\pi^2+1)b_n^2. \end{eqnarray} I believe that $$ \sum_{n=0}^\infty(n^2\pi^2+1)a_n^2-(\sum_{n=0}^\infty a_n)^2\ge 0$$ and now I do not have time to prove it. So $$ \int_0^1(f(x))^2dx+\int_0^1(f'(x))^2dx\ge\frac12(f(0))^2. $$

Answer 2

One can do as follows. We may assume that $f(0)=1$. Let $$ J(f) = \int_0^1 (f'(x)^2 + f(x)^2) dx \geq 0. $$ We want to minimize $J(f)$ under the condition that $f(0)=1$. Let $F(f',f) = f'^2+f^2$. From variational calculus we know that any minimum satisfies $$ F_f' - \frac{d}{dx} F_{f'}' = 0 $$ and $F_{f'}'(f'(1),f(1))=0$. This simplifies to solving $$ f'' = f $$ with the boundary conditions $f(0)=1$ and $f'(1)=0$. The solution is $$ f(x) = -\tanh(1)\sinh(x) + \cosh(x). $$ One then checks that for this $f$ we have that $$ J(f) > \frac{1}{2} = \frac{1}{2} f(0)^2. $$ This proves the inequality.

(In fact, we have $J(f) = \tanh 1 > 1/2$.)

Answer 3

Here is a variant which will yield an improved, but not optimal, constant in the inequality.

Let $\lambda$ be a positive real number whose exact value will be specified later. Application of the fundamental theorem of calculus and the arithmetic-geometric inequality gives \begin{eqnarray} f(0)^2 &=& f(x)^2 - \int_0^x 2f(\xi)f'(\xi)\,d\xi \nonumber\\ &\leq& f(x)^2 + \int_0^x \frac{1}{\lambda}f(\xi)^2 + \lambda f'(\xi)^2\, d\xi \nonumber\\ &\leq& f(x)^2 + \int_0^1 \frac{1}{\lambda}f(\xi)^2 + \lambda f'(\xi)^2\, d\xi. \end{eqnarray} If we integrate both sides of this inequality from $0$ to $1$ and, moreover, choose $\lambda$ such that $1+1/\lambda=\lambda$, that is $\lambda = (1+\sqrt{5})/2$, then we arrive at \begin{equation} \frac{2}{1+\sqrt{5}}f(0)^2 \leq \int_0^1 f(x)^2 + f'(x)^2 \, dx. \end{equation} As $2/(1+\sqrt{5}) \approx 0,6180 > 1/2$, the desired inequality follows.