Let $f:[0,2] \rightarrow \mathbb{R}$ continuous and positive such that $\int_0^1 f(x) \, dx=\int_1^2 f(x) \, dx=1$ for each $x \in [0,1]$ prove that there is a unique $g(x) \in[1,2]$ such that $\int_x^{g(x)} f(t) \, dt=1$ prove that the function $g:[0,1] \rightarrow \mathbb{R}$ is the class $\mathbb{C}^1$

by the fundamental theorem of calculus

$$ F(1)-F(0)= \int_0^1 f(x) \, dx= \int_1^2 f(x)=F(2)-F(1)=1$$ then $2F(1)=F(2)+F(0)$.

  • $\begingroup$ Consider the function $F(y) = \int_x^{y} f(t)dt$ for any fixed $x\in[0,1]$. What can you say about $F(y=x)$ and what can you say about $F(y=2)$? You want to compare these two values to $1$ and apply the intermediate value theorem. $\endgroup$ – Winther Sep 16 at 23:51

If $x \in [0, 1]$ and $g(x) \in [1, 2]$, then $$\int_x^{g(x)}f(t)\,dt = \int_x^1 f(t)\,dt + \int_1^2 f(t)\,dt -\int_{g(x)}^2 f(t)\,dt = 1$$

This simplifies to showing that $$\int_x^1 f(t)\,dt = \int_{g(x)}^2 f(t)\,dt \tag 1$$ has a solution $g(x) \in [1, 2]$. Since the LHS ranges over $[0, 1]$, it is sufficient to show that a $g(x)$ exists such that $\int_{g(x)}^2 f(t) \, dt$ ranges over $[0, 1]$. This is easy to show using the intermediate value theorem since $\int_1^2 f(t) \, dt = 1$, $\int_2^2 f(t)\,dt = 0$, and $f(x)$ is continuous.

Now for continuity and differentiability: Let $F(x) = \int_0^x f(t)dt$. Then you are trying to show that there exists a continuous and differentiable $g(x) \in [1, 2]$ such that $$F(g(x))-F(x)=1$$

The solution in $g(x)$ would be $$g(x) = F^{-1}(1+F(x))$$ where $F^{-1}(x)$ denotes the inverse of $F(x)$. Note that $F(x)$ is strictly increasing (since $f(x)>0$) so $F^{-1}(x)$ would also be continuous. Then since the composite of two continuous functions is also continuous, $g(x)$ would also be continuous. The derivative of $g(x)$ would be $$g'(x) = \frac{f(x)}{f(F^{-1}(1+F(x)))} = \frac{f(x)}{f(g(x))}$$

which would be defined for $x \in [0, 1]$.

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  • $\begingroup$ Please, could you clarify the LHS ranges over $[0,1]$, it is sufficient to show that a continuous $g(x)$ exists such that $\int_{g(x)}^{2} f(t) dt$ ranges over $[0,1]$. what does it mean LHS ranges? $\endgroup$ – Mayra Isabel Ferreira Ortiz Sep 16 at 21:58
  • $\begingroup$ @MayraIsabelFerreiraOrtiz I was saying that the left hand side (LHS) of $(1)$ would be in $[0, 1]$. $\endgroup$ – Varun Vejalla Sep 16 at 22:06
  • $\begingroup$ why is it easy to show that $g$ exists, I still don't understand why using $\int_{1}^{2} f(t)dt=1, \int_{2}^{2} f(t) dt=0$ and $f(x)$ is continuous, Please could you explain it to me? $\endgroup$ – Mayra Isabel Ferreira Ortiz Sep 16 at 22:13
  • $\begingroup$ Let $h(x)=\int_x^2 f(t) dt$. Then $h(x)$ must be continuous since the integral of a continuous function would also be continuous and since $h(1) = 1$ and $h(2) = 0$, by the intermediate value theorem, $h(x) = y$ would have some solution $x$ for all $y \in [0, 1]$ $\endgroup$ – Varun Vejalla Sep 17 at 0:48
  • $\begingroup$ It is not so clear why $g$ is continuous and differentiable. Can you include the proof? $\endgroup$ – Arctic Char Sep 17 at 3:54

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