# a Differential inequality without integration

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?

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.

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.