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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ a differentiable function with $f(x)+f^{'}(x)\leq1$ for all $x \in \mathbb{R}$ and $f(0)=0$. Which is the maximum possible value of $f(1)$?

The question is "solved" here : maximum value and a differential inequality

but I did a mistake by supposing that $\frac{d}{d x} (e^x f(x)) $ is integrable in the interval $[0,1]$. I don't know how to solve the problem without such condition. Someone could help me?

thanks in advance

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The solution to $f'(x) + f(x) = 1$ with $f(0) = 0$ is given by $f(x) = 1 - e^{-x}$ as can be shown via integrating factors for example.

Inspired by this integrating factor method, if $f'(x) + f(x) \leq 1$, you can let $g(x) = e^x f(x) - (e^x - 1)$, and then you have $g'(x) \leq 0$ with $g(0) = 0$. So $g(x)$ is decreasing with $g(0) \leq 0$. Hence $g(1) \leq 0$.

Writing this out, we see that $ef(1) - (e - 1) \leq 0$ or $f(1) \leq 1 - e^{-1}$. The function $f(x) = 1 - e^{-x}$ from above will achieve this minimum.

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Assume $f(\alpha)=a,f(\beta)=b,\alpha<\beta$ and $\forall x\ge\alpha,f(x)\ge f(\alpha)=a$.

Then, $\forall x\ge\alpha,f'(x)\le 1-a\Rightarrow b\le a+(\beta-\alpha)(1-a)$.

Assume there exists $f$ satisfying all the conditions of the question with $f(1)=M$.

Fix positive integer $N$. Since $f$ is continuous on $[0,1]$, we have $l_n:=\inf\{x\in[0,1]\mid f(x)=\frac{n}{N}\cdot M\}$ for $n=1,...,N$.

Then, $\forall n,\frac{M}{N}\le(l_{n+1}-l_n)(1-\frac{n}{N}\cdot M)\Rightarrow\frac{1}{N}\cdot\frac{M}{1-\frac{n}{N}M}\le l_{n+1}-l_n$.

Therefore, $1\ge l_N\ge M\cdot\frac{1}{N}\sum_{n=1}^N\frac{1}{1-M\cdot\frac{n}{N}}$.

Limiting $N$ to $\infty$, $1\ge \int_0^1\frac{Mdx}{1-Mx}=\int_0^M\frac{dy}{1-y}=-\ln(1-M)$.

Therefore, $M\le 1-e^{-1}$.

Now you have natural equality condition by solving $f+f'=1$ and this shows that $1-e^{-1}$ is actually obtainable.

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