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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 $3e^x = \int_0^3 e^x dt$?2012-12-06
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    But $f$ must be differentiable in order for a multiple of $f$ to equal that integral since you assumed $f$ to be continuous.2012-12-06
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    @Tobias: So why is that?2012-12-06
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    The fundamental theorem you yourself mentioned says that if $f$ is continuous then that integral is differentiable.2012-12-06
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    If $f$ is continuous then the primitive of $f$, $F$ is differentiable, but that does not imply that $f$ is differentiable, right?2012-12-06
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    Correct. But you are also assuming that $f$ equals a multiple of its own integral, which is what then makes $f$ differentiable. Note that if you change the lower bound on the integral to $-\infty$ then such $f$ do exist, namely the exponential functions.2012-12-06
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    Ok, I see, so I guess I have a proof by contradiction that such a function does not exist?2012-12-06
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    The function whose value is everywhere $0$ satisfies this equation.2012-12-06
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    @MichaelHardy: Yes, therefore I assumed $f \neq 0$.2012-12-06
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    ...and if $\lambda f(s) = \int_0^s(\text{some continuous function})\,dt$ then that implies $f$ is differentiable.2012-12-06
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    @Imray : Just try to evaluate for any $x$ : your example doesn't work.2012-12-06
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    @PatrickDaSilva Actually the problem seems to be that Imray is taking $s$ to be the constant $3$, which is doubly strange considering that $f$ is only defined on $[0,1]$.2012-12-06
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    No I was taking $f$ to be $e^x$ and $\lambda$ to be 3. I thought that worked but I guess I didn't understand the question.2012-12-06
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    @Imray Your right-hand side clearly assumes that $s=3$. The relation is supposed to hold for all $0 \le s \le 1$ (it is not stated explicitly so the natural convention is to require it for all applicable $s$).2012-12-06
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    @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

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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?2012-12-06
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    @Imray Obviously your question has nothing to do with the question we are discussing!2012-12-06
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    Can you please explain why the given condition imply that f is $ C^1$2018-10-02
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    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
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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 $t$.