Is it true that if $f\in W^{1,p}(\mathbb{T})$ then the difference quotient of $f$ converges in $L^p(\mathbb{T})$ to the weak derivative of $f$? Let $\mathbb{T}$ be the 1-torus and if $1\le p \le+\infty$ define the Sobolev space $$W^{1,p}(\mathbb{T}):=\left\{f\in L^p(\mathbb{T})\ |\ \exists g\in L^p(\mathbb{T}), \forall\varphi\in C^\infty(\mathbb{T}), \int_{-\pi}^\pi f(t)\varphi'(t)\operatorname{d}t=-\int_{-\pi}^\pi g(t)\varphi(t)\operatorname{d}t\right\}.$$
It is known that if $f\in W^{1,p}(\mathbb{T})$ then the function $g$ whose existence is stated in the previous definition is (a.e.) unique and we'll refer to this function with $D_w(f)$.

Is is true that $$\forall f\in W^{1,p}(\mathbb{T}), \left\| \frac{f(\cdot+h)-f}{h} -D_w(f)\right\|_p\rightarrow 0, h\rightarrow0?$$

I proved that the claim is true for the value $p=2$ using the Fourier transform and that for $p=\infty$ the claim is false (see the comments) but it is true if $W^{1,\infty}(\mathbb{T})$ is replaced by $C^1(\mathbb{T})$. Also, I know that $\frac{f(\cdot+h)-f}{h}$ weak converges in $L^p(\mathbb{T})$ to $D_w(f)$ if $1<p<\infty$, weak* converges in $L^p(\mathbb{T})$ to $D_w(f)$ if $p=\infty$, weak* converges in the complex measures to $D_w(f)$ if $p=1$ and, finally, I know that $\frac{f(\cdot+h)-f}{h}$ converges pointwise $\mu$-a.e. to $D_w(f)$ if $p>1$ (see Evans, Gariepy - Measure theory and fine properties of functions, chapters 4 and 6).
However, I don't know if the claim is true for the other values of $p$ and if the cited results could help somehow to prove it.
 A: The question has a positive answer for $1\le p<\infty$ as pointed out by reuns in the comments. The details follow.
First, notice that if $f\in W^{1,p}(\mathbb{T})$ then $f\in W^{1,1}(\mathbb{T})$ and so, after changing $f$ on a set of zero measure, we can assume that $f$ is absolutely continuous with derivatives equals a.e. to $D_w(f)$. So, if $0<h<\pi$:
$$f(x+h)-f(x)=\int_{x}^{x+h}D_w(f)(t)\operatorname{d}t\\=2\pi\int_{x-\pi}^{x+\pi}\chi_{[-h,0]}(x-t)D_w(f)(t)\frac{\operatorname{d}t}{2\pi}=2\pi\int_{-\pi}^{\pi}\chi_{[-h,0]}(s)D_w(f)(x-s)\frac{\operatorname{d}s}{2\pi}=(2\pi\chi_{[-h,0]}*D_w(f))(x).$$
Then:
$$\frac{f(\cdot+h)-f}{h}=\frac{2\pi\chi_{[-h,0]}}{h}*D_w(f).$$
Now, notice that:
$$\left(\frac{2\pi\chi_{[-h,0]}}{h}\right)_{h\in (0,\pi)}$$
is a summability kernel for $h\rightarrow0^+$. Then:
$$\left\|\frac{f(\cdot+h)-f}{h}-D_w(f)\right\|_p=\left\|\frac{2\pi\chi_{[-h,0]}}{h}*D_w(f)-D_w(f)\right\|_p\rightarrow0, h\rightarrow 0^+.$$
Analogously we can reason for $-\pi<h<0$, and so the claim follows.
