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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 $[0,1]$.

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What about $3e^x = \int_0^3 e^x dt$? – CodyBugstein Dec 6 '12 at 19:31
But $f$ must be differentiable in order for a multiple of $f$ to equal that integral since you assumed $f$ to be continuous. – Tobias Kildetoft Dec 6 '12 at 19:33
The function whose value is everywhere $0$ satisfies this equation. – Michael Hardy Dec 6 '12 at 19:48
@MichaelHardy: Yes, therefore I assumed $f \neq 0$. – Yiteng Dec 6 '12 at 19:52
@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. – Patrick Da Silva Dec 6 '12 at 22:50

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.

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Why won't $3e^x = \int_0^3 e^x dt$ work? – CodyBugstein Dec 6 '12 at 19:56
@Imray Obviously your question has nothing to do with the question we are discussing! – Mercy King Dec 6 '12 at 20:00

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 $t$.

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