If $f:[0,1]\to\mathbb{R}$ is an integrable function, prove the following:

Question

If $f:[0,1]\to\mathbb{R}$ is an integrable function, prove the following:

$\displaystyle\lim_{h\to0}\int_{[0,1]}\frac{|1+h\cdot f(t)|-1}{h}dm(t)=\int_{[0,1]}f(t)dm(t)$

I don't even know where to get started on this one. When I saw the absolute values, my first thought was to use the triangle inequality somewhere...but the problem is an equality rather than an inequality, which means I would have to show both directions, and the "backward" direction seems iffy.

My next thought was to use some theorem that can pull the limit inside the integral. I can't remember all of them, but do recall the Monotone Convergence Theorem, the Lebesgue Dominated Convergence Theorem (LDCT) and the Generalized LDCT, but none of these seem particularly useful in this case. Any insight would be much appreciated! Thanks in advance!

Answer 1

Here's a hint: $\bigl||1+u|-1\bigr|\le|u|$.

Answer 2

For any $t\in [0,1]$ you have $$\lim_{h\to 0} \frac{\vert 1+hf(t)\vert-1}{h}=f(t)\, . $$ Indeed, this is obvious if $f(t)=0$; and if $f(t)\neq 0$ you can write the quotient as $$f(t)\times \frac{\vert 1+hf(t)\vert-1}{hf(t)} $$ and use the fact that the function $u\mapsto \vert u\vert$ is differentiable at $u=1$ with derivative equal to $1$.

Now, you can use the dominated convergence theorem because
$$\left\vert \frac{\vert 1+hf(t)\vert-1}{h}\right\vert\leq \vert f(t)\vert$$ for all $h$ and every $t\in [0,1]$, by the triangle inequality.

Answer 3

This is equivalent to

$$\lim_{h\to0}\int_{[0,1]}\left(\frac{|1+h\cdot f(t)|-1}{h} - f(t) \right) dm(t) = 0$$

A common method for problems like this is that you know that one part of the expression is approximately equal to something simple, so you split it apart into the sum of that something plus the error (or the product of that something times the relative error, or other similar things). It's often easier to deal with the approximation part of the problem if it's approximating zero rather than something more complicated.