# Differentiability for an integral function

Mathematics Asked by andereBen on December 4, 2020

I can’t conclude the following exercise:

Let $$f in L^{1}[0,2]$$ , $$psi:[0,1] rightarrow [0,1]$$ a function and $$F[0,1] rightarrow mathbb{R}$$ defined by $$F(t)= int_0^1 f(x+psi(t))dx$$ for every $$t in [0,1]$$.

• Show that if $$psi$$ is continuous then $$F$$ is continuous.
• Show that if $$psi in C^{1}$$ and $$psi'(t)>0$$ for every $$t in [0,1]$$, then $$F$$ is differentiable for almost every $$t in [0,1]$$.

• Let $$s: I consider the difference

$$|F(t)-F(s)| = left| int_0^1Bigl[f(x+psi(t)) – f(x+psi(s)) Bigr] dx right| leq int_0^1 left| f(x+ psi(t)) – f(x+psi(s)) right|dx$$

Taking the limit as $$s rightarrow t^{-}$$ and using the fact that $$f in L^{1}$$, I can pass the limit inside the sign of integral in the rhs, and I obtain the continuity.

• Here if $$f$$ would be differentiable, everything would work. Since it is only $$L^1$$, this function can’t be differentiated, so I was thinking to split the variables in $$f(x+psi(t))$$, but I don’t know how.

Any help/answer/hint is highly appreciated

As pointed out by @Kavi Rama Murthy, your proof of continuity is not correct as one cannot deduce the a.e.-pointwise convergence of the integrand.

(a) For the continuity, recast the integral as

$$F(t) = int_{0}^{2} f(x)mathbf{1}_{[psi(t),psi(t)+1]}(x) , mathrm{d}x$$

and then note that now one can apply the dominated convergence theorem.

(b) The above representation also allows to solve the second part. To this end, let

$$E = { u in [0, 1] : text{both u and u+1 are Lebesgue points of f} }.$$

If $$Z$$ denotes the complement of the set of Lebesgue points of $$f$$ in $$[0, 2]$$, then $$Z$$ has measure zero by the Lebesgue differentiation theorem. Moreover,

$$[0,1]setminus E subseteq Z cup (Z-1),$$

and so, $$E$$ has full measure in $$[0,1]$$. Now by the assumption, $$psi$$ has a $$C^1$$ inverse, and so, $$psi^{-1}(E)$$ also has full measure in $$[0,1]$$. Finally, if $$t in psi^{-1}(E)$$, then both $$psi(t)$$ and $$psi(t)+1$$ are Lebesgue points of $$f$$, and so,

$$frac{F(t+h) - F(t)}{h} = frac{1}{h} int_{psi(t)+1}^{psi(t+h)+1} f(x) , mathrm{d}x - frac{1}{h} int_{psi(t)}^{psi(t+h)} f(x) , mathrm{d}x$$

converges as $$h to 0$$. This completes the proof.

Correct answer by Sangchul Lee on December 4, 2020

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