The convergence in sense of distribution associated with Lebesgue integrable function but not differential.

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Let $f \in L^1(\mathbb{R}^n)$ and $T$ is a distribution. Show that $T_{D_j^hf} \rightarrow \partial x_j T_f$ as $h \rightarrow 0$ in the sense of distributions, where $$D_j^hf(x) := \dfrac{f(x + e_jh) − f(x)}{ h}, \quad h>0.$$

I try to prove that, for any $\phi \in C^{\infty}_c(\mathbb{R}^n$
$lim_{h \rightarrow 0} \int_{\mathbb{R}^n}\phi(x).\dfrac{f(x + e_ih) − f(x)}{ h}dx=-\int_{\mathbb{R^n}}\partial x_i\phi(x).f(x)dx.$

I am planning to bounded the integral

$$\int_{\mathbb{R}^n}\left\vert \phi(x).\dfrac{f(x + e_ih) − f(x)}{ h}dx + \partial x_i\phi(x).f(x)\right\vert dx < \varepsilon$$ for all $|h|<\delta.$

But I don't know how to deal with $\dfrac{f(x + e_ih) − f(x)}{ h}$ in this case. Could you please give me some ideas?

Btw, can you recommend me some good books about Distribution Theory?

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As usual in the theory of distributions we want to "throw" everything onto the test function. After a change of variables (shifting the first term) we have

\begin{align*} \int_{\mathbb{R}^d} \phi(x)D_j^h(x) dx&=\int_{\mathbb{R}^d} \frac{\phi(x)}{h} f(x+he_j)dx +\int_{\mathbb{R}^d} \frac{\phi(x)}{h}f(x)dx\\ &=\int_{\mathbb{R}^d} \frac{\phi(x-he_j)-\phi(x)}{h} f(x)dx. \end{align*} Now you can us dominated convergence to conclude (as the factor with $\phi$ has compact support and is bounded by the mean value theorem).