While working on a certain problem in complex analysis - and specifically in finding harmonic conjugates, I needed to justify a certain equality:
Suppose $f \in C^1(\mathbb{R})$ and $c > 0$. Then: $$\lim_{h \to0} \int^{c}_{0}\frac{f(t+h)-f(t)}{h}dt = \int^{c}_{0}\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}dt=\int^{c}_{0}f'(t)dt$$
My proof:
We will prove this by using Heine definition of the limit: let $(h_n)_{n=0}^{\infty}$ be a sequence s.t $\lim_{n\to \infty}h_n = 0$. So we need to prove: $$\lim_{n \to\infty} \int^{c}_{0}\frac{f(t+h_n)-f(t)}{h_n}dt = \int^{c}_{0}f'(t)dt$$ Let us define a function series $g_n:\mathbb{R}\to\mathbb{R}$ s.t $\forall n\in \mathbb{N}:\;g_n(x) = \frac{f(x+h_n)-f(x)}{h_n}$. We will prove uniform convergence $g_n\to f'$ in $[0,c]$. Let us remember that $f \in C^1(\mathbb{R})$, so $f'$ is uniformly continuous in $[-1,c+1]$.
Let there be $\varepsilon > 0$ and $x\in [0,c]$, from uniform continuity there exists a $\delta>0\;s.t\;\;\forall x,y\in [-1,c+1]:\;|x-y|<\delta \implies |f'(x) - f'(y)|< \varepsilon$.
From Lagrange's mean value theorem: There exists a $0<\theta_n<1$ s.t: $$\frac{f(x+h_n)-f(x)}{h_n}=f'(x+\theta_nh_n)$$ Let's remeber that $h_n\to0$, so there exists $n_0$ s.t $\forall n>n_0:\;h_n< \min\{1,\delta\}$ and then: $$|g_n(x)-f'(x)| = |f'(x+\theta_nh_n)-f'(x)|<\varepsilon$$ Thus $g_n$ converges uniformly to $f'$ in $[0,c]$ and this concludes the proof - because we can exchange limit and integral.
Is this proof correct? Because I've come across several people who though that we need to enforce $f\in C^2(\mathbb{R})$ for this conclusion to be correct but I don't think it is necessary.
I would really appreciate any help!
This looks absolutely correct. You certainly do not need $C^2$. An alternative reasoning is to note that since $f'$ is continuous on $[-1, c+1]$, then $f$ is Lipschitz continuous on that very interval. Thus for $h \in (-1, 1) \setminus \lbrace 0 \rbrace$ and $t \in [0, c]$ there exists $L>0$: $$ \left \lvert \frac{f(t+h)-f(t)}{h} \right \rvert \leq L \left \lvert \frac{h}{h} \right \rvert = L $$ So you can invoke dominated convergence theorem to conclude that limit and integral are interchangeable.