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Let $\phi, a \in C^{\infty}([0,1])$ and assume $a(0)=1$. Suppose that $$ \int_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt =0 \qquad \text{for all $\tau \in \mathbb R$}. $$ Does it follow that $\phi$ is a constant in a neighborhood of $t=0$?

If the answer to the above question is affirmative I have a follow up question as follows: Suppose additionally that $b \in C^{\infty}([0,1]$ with $b(0)=1$ and instead of the latter equation there holds $$\Bigg |\int_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt\Bigg| \leq \frac{1}{|\tau|} \Bigg|\int_0^1 e^{\tau \,\phi(t)}\,b(t)\,dt\Bigg|$$ Does it again follow that $\phi(t)$ is constant in a neighborhood of $t=0$?

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    $\begingroup$ no; just try $\phi(t)=\text{constant}\neq 0$ and take for $a(t)$ any function with $\int_0^1 a(t)dt=0$; then obviously also $e^\tau \int_0^1 a(t)dt=0$ for any $\tau\in\mathbb{R}$. $\endgroup$
    – Carlo Beenakker
    Commented Mar 14, 2022 at 16:44
  • $\begingroup$ This was a bad typo on my part. I meant to write derivative of $\phi$ in the question. sorry about this. $\endgroup$
    – Ali
    Commented Mar 14, 2022 at 17:00

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No. E.g., let $\phi(t)=(t-1/2)^2$ (so that $\phi$ is even about $1/2$) and $a(t)=2(1/2-t)$ (so that $a$ is odd about $1/2$).

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