Integral limit $\lim _{t\to 0}\frac 1 t \int_0^1 (f(x+t)-f(x))x \, dx$

Question

Let $f:\left[0,1\right]\rightarrow\mathbb{R}$ be a continuous function and $\int_0^1 f(x) \, dx=0$. Compute the following limit: $$\lim _{t\to 0} \frac 1 t \int_0^1 (f(x+t)-f(x))x \, dx$$ I thought about Lagrange's Theorem but we don't know if $f$ is differentiable so it can't be used in this case.

Answer 1

For the problem of the definition of $\int_0^1\left(f\left(x+t\right)-f\left(x\right)\right)xdx$ (since $f$ is defined on $[0,1]$,the latter integral only has sense for $t=0$), see my comment under the original question.

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In the following I will suppose that is not an issue (It is sufficient for example to assume that $f$ is defined and continuous over $[0,1+\varepsilon)$ with $\varepsilon>0$)

Act as if $f$ was $\mathcal{C}^1$ at first (the inversion limit/integral is then alright from domination convergence theorem) to have an idea of what the limit can be : $$ \lim _{t\to 0}\frac{1}{t}\int_0^1\left(f\left(x+t\right)-f\left(x\right)\right)xdx = \int_0^1 f'(x)xdx= [f(u)u]_0^1 -\underbrace{\int_0^1 f(x)dx}_{=0}=f(1). $$

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In the general case, let $$ F(t):= \int_0^1 f\left(x+t\right) xdx$$ then the limit you seek is $$\lim _{t\to 0}\frac{F(t)-F(0)}{t} = F'(0)$$

Now \begin{align*} F(t) & = \int_0^1 f\left(x+t\right) xdx \\ & = \int_t^{1+t}f(y)(y-t) dy \tag{$y=x+t$}\\ & = \int_t^{1+t}f(y)y dy -t\int_t^{1+t}f(y) dt \end{align*} So (since if $G(t):=\int_0^t h(y) d y$ with $h$ continuous, $G$ is differentiable and $G'(t)=h(t)$) : $$F'(t) = f(1+t)(1+t)-f(t)t-t(f(1+t)-f(t)) -\int_t^{1+t}f(y) dy$$ and $$ F'(0)=f(1)- \int_0^{1}f(y) dy=f(1).$$ So the limit remains true in the general case.