2
$\begingroup$

Let $\lambda \in \mathbb{R}$ be a constant, $\lambda \neq 0$. Is there a function $f \in C[0,1]$, $f \neq 0$, that satisfies the following relation:

$\lambda f(s) = \int_0^s f(t) \, dt$

Attempt at a solution:

Applying Fundamental Theorem of Calculus, one gets $ \lambda f(s) = F(s) - F(0)$, which implies that $f(0) = 0$. It follows that $f$ cannot be differentiable, because if it were, then taking the dervative on both sides gives $\lambda f'(s) = f(s)$, which gives together with the boundary condition $f(s) = \exp[\frac{s}{\lambda}] - 1$, but this function does not satisfy the relation. Hence if such a function exists, then it cannot be differentiable on $[0,1]$.

  • 1
    @Imray : it's not because you didn't understand the question, it's because you wrote something that didn't make sense. $3 e^s = \int_0^s e^x \, dx$ is false, even if you did a typo ; in other words, $f(x) = e^x$ doesn't work here. It's as simple as : $ \int_0^s e^x dx = e^s - 1 \neq \lambda e^x $ whatever the value of $\lambda$ you choose. – 2012-12-06

2 Answers 2

3

Since $f$ is continuous, it follows from the given relation that it is $C^1$ and solves the DE $ y'=\lambda^{-1}y, $ i.e. $ f(t)=c\exp(\lambda^{-1}t) \quad t \in [0,1], $ where $c$ is a constant. Since $c=f(0)=0$, we have $f \equiv 0$. Hence there is no such function.

  • 0
    Since $f\in C[0,1]$, and $f’=\lambda^{-1}f \in C[0,1]$ it follows that $f \in C^{1}[0,1]$ – 2018-10-16
3

Let $M(r) = \max \{|f(x)|: 0 \le x \le r\}$. Then $|\lambda| |f(x)| \le M(r) x \le M(r) r$ for $0 \le x \le r$, so $|\lambda| M(r) \le M(r) r$. Thus if $r < \lambda$ we have $M(r)=0$, i.e. $f(x) = 0$ for $0 \le x < |\lambda|$.

Let $t$ be the greatest $r$ such that $M(r) = 0$. I claim $t = 1$. If not, for $t \le x \le t+r$ (where $t + r \le 1$) we have $|\lambda| |f(x)| = \left|\int_t^x f(x)\ dx \right| \le M(t+r) r$, so $|\lambda| M(t+r) \le M(t+r) r$, and if $r < |\lambda|$ we again get $M(t+r) = 0$, contradicting the maximality of $t$.