Is there a function $f \in C[0,1]$ such that $f(0)=f(1)=0$, $f'(0)=0$, $f'(1)=1$ and $\|f\|_{C[0,1]} < \epsilon$, $\|f'\|_{C[0,1]} < \epsilon$

Question

Given $\epsilon \in (0,1)$ I am looking for a twice continuously differentiable function $f\colon [0,1] \to \mathbb{R}$ satisfying the following conditions:

• $$ f(0)=f(1) =f'(1)=0, \quad f'(0)=1, $$
• $$ \sup_{x \in [0,1]} |f(x)| \leq \epsilon, \qquad \int_0^1 |f'(x)| dx \leq \epsilon. $$

It is quite obvious that such function exists, since we can set $f(x) = 0$ for $x \in [\epsilon,1]$, and on the interval $[0,\epsilon]$, $f$ can resemble $$1_{[0,\epsilon]}(x)x + 1_{[\epsilon,2\epsilon]}(x)(2\epsilon - x).$$ Of course the above function is not (even once) differentiable, so I would like a smooth version of it. Is there some simple formula for such function?

Answer 1

You can simply take $\begin{cases}f_n(x)=(1-x)^nx\\f_n'(x)=(1-x)^n-n(1-x)^{n-1}x\end{cases}$

$f_n$ is a polynomial thus it has all regularity you can hope for.

$f_n(0)=f_n(1)=0$ as long as $n>0$

$f_n'(0)=1$ and $f_n'(1)=0$ as long as $n>1$

The maximum is reached in $x_0=\frac 1{n+1}$.

$||f_n||_\infty=f_n(x_0)=\dfrac{(1-\frac 1{n+1})^n}{n+1}\sim \dfrac 1{ne}<\epsilon\quad$ for $n$ large enough.

$\displaystyle||f_n'||_1=\int_0^{x_0}f_n'(x)\mathop{dx}-\int_{x_0}^{1}f_n'(x)\mathop{dx}=\dfrac{2\ n^n}{(n+1)^{n+1}}\sim \dfrac 2{en}<\epsilon\quad$ for $n$ large enough.

The family of functions $C_{p,q}\, x^p(1-x)^q$ in $[0,1]$ is generally usefull when you consider examples involving $||\cdot||_\infty$ and $||\cdot||_1$ norms.

Answer 2

After some thinking the function $$ f_n(x) = \frac{1}{\pi} \sin(\pi x) \exp\left( -\frac{nx}{(1-x)} \right) $$ should work for sufficiently large $n$.