distribution - exercises Let $f \in \mathcal{C}(\mathbb{R})  \cap L^1(\mathbb{R}).$ We suppose that $f$ is bounded and even $(f(-x)=f(x))$. let $\lambda \in \mathbb{R}.$
We assume that: 
$\int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} dx = \lim_{\varepsilon \to 0} \int_{\varepsilon < |x| < 1} \dfrac{1 - f(\lambda x)}{x} dx= 0 $
and forall $\varphi \in \mathcal{D}(\mathbb{R}),$ we have: 
$$\int_{-\infty}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx = \int_1^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx + \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(0)) dx$$
and 
$< T_{\lambda} , \varphi > = \int_{\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx$ for all $\varphi \in \mathcal{D}(\mathbb{R}.$ We have 
$\forall \varphi \in \mathcal{D}(\mathbb{R}): |\int_1^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| \leq C N_1(\varphi).$
The question is: 
1- prove that 
$$\exists C > 0, \forall \varphi \in \mathcal{D}(\mathbb{R}): |<T_{\lambda} , \varphi >| \leq C N_1(\varphi)|.$$
My solution is: 
$|<T_{\lambda} , \varphi > | = |\int_{-\infty}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx| = |\int_{1}^{+ \infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| + \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(0)) dx$
$\leq |\int_{1}^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| + \sup_{[-1,1]} |\varphi(x) - \varphi(0)| \int_{-1}^1 \dfrac{1 - f(\lambda x)}{x} dx|$
$\leq C N_1(\varphi)$
my solution is correct?
EDIT: i have an other question.
I have an other question please.
2- Prove that $$\forall \varphi \in \mathcal{D}(\mathbb{R}), \exists M > 0, \forall n \in \mathbb {N}: |<T_n - vp \dfrac{1}{x} , \varphi >| \leq M \displaystyle\int_{-\infty}^{+\infty} |f(nx)| dx$$ use to this proof that $< vp \dfrac{1}{x},\varphi> = \displaystyle\int_0^{+\infty} \dfrac{\varphi(x) - \varphi(-x)}{x} dx$
my solution is:$|<T_n - vp \dfrac{1}{x},\varphi>| = |\int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx - \int_0^{+\infty} \dfrac{\varphi(x) - \varphi(-x)}{x} dx| = |\int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) - \int_0^{+\infty} \dfrac{\varphi(x)}{x} dx - \int_{-\infty}^0 \dfrac{\varphi(y)}{y} dx| = |\int_{- \infty}^{+ \infty} - \dfrac{f(\lambda x)}{x} \varphi(x) dx| \leq \int_{-\infty}^{+ \infty} |f(nx)| dx / \sup_{x \in K} \dfrac{|\varphi(x)|}{x} dx$ where $K$ is support of $\varphi,$ then $ |<T_n - vp \dfrac{1}{x},\varphi>| \leq M \int_{-\infty}^{+\infty} |f(nx)|dx$ . My solution is correct? ?
QUESTION 3.
3- Deduce that $T_n \longrightarrow vp \dfrac{1}{x}$ in $S'(\mathbb{R})$ when $n \rightarrow + \infty.$
My solution is, 
we have $\forall \varphi \in \mathcal{D}(\mathbb{R}), |< T_n - vp \dfrac{1}{x} , \varphi>| \leq M \int_{-\infty}{+\infty} |f(nx)|dx$
so $f(nx)$ converge almost every where to 0 when $n \rightarrow + \infty$ because $f(nx)$ is bounded and $f \in L^1,$ then by Lebesgue theorem (dominated convergence), $\int_{-\infty}^{+\infty} |f(nx)| dx \rightarrow 0$ when $n \rightarrow + \infty$
then, $<T_n , \varphi > \longrightarrow < vp \dfrac{1}{x},\varphi>$ so $T_n \longrightarrow vp \dfrac{1}{x}$ in $S'.$
QUESTION: (EDIT)
Let $\forall \varphi \in \mathcal{D}(\mathbb{R}): < T_{\lambda},\varphi> = \int_{-\infty}^{+\infty} \dfrac{1 - f(\lambda x)}{x} \varphi(x) dx$ where $\lambda \in \mathbb{R}.$ Prouve that $$\forall \varphi \in \mathcal{D}(\mathbb{R}): |\int_1^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx|\leq C N_1(\varphi)$$
my question is, if we can write this solution; $|\int_1^{+\infty} \dfrac{1 - f(\lambda x)}{x} (\varphi(x) - \varphi(-x)) dx| = \int_K |\dfrac{1 - f(\lambda x)}{x}||\varphi(x) - \varphi(-x)| dx \leq \int_K |\dfrac{1 - f(\lambda x)}{x} 2x |\varphi'(\xi_x)| dx \leq C \sum_{x \in K} |\varphi'(x)| \leq C N_1 (\varphi).$
They are errors in this solution please?
 A: Question 1: You have
$$
|<T_\lambda, \varphi>| \leq CN_1(\varphi) + \int_{-1}^1 \Big|\frac{1-f(\lambda x)}{x}(\varphi(x)-\varphi(0))\Big| dx. $$
By Taylor, $\varphi(x)-\varphi(0) = x\varphi'(\xi_x)$, so the integral becomes
$$
\int_{-1}^1 \Big|1-f(\lambda x)\Big| |\varphi'(\xi_x)| dx \leq \int_{-1}^1 |1-f(\lambda x)| dx \sup_{\xi\in[-1,1]} |\varphi'(\xi)|  \leq C' N_1(\varphi).$$
Question 2: You have (see your proof) - treating $\int_{-\infty}^\infty \frac{\dots}{x} dx$ as $\lim_{\epsilon\to 0} \int_{\mathbb{R}\setminus [-\epsilon,\epsilon]} \frac{\dots}{x} dx$ - 
$$ |<T_n - vp\frac{1}{x},\varphi>| \leq \int_{-\infty}^\infty \Big| \frac{f(nx)}{x} \varphi(x)\Big| dx \leq  \int_{-\infty}^\infty \Big|\frac{f(nx)}{x}\varphi(0)\Big| dx + \int_{-\infty}^\infty |f(nx)\varphi'(\xi_x)| dx $$
again using Taylor's $\varphi(x)=\varphi(0)+x\varphi'(\xi_x)$. The first summand is $0$ as
$f$ is even and so you conclude with $M:=N_1(\varphi)$ for all $n$
$$ |<T_n - vp\frac{1}{x},\varphi>| \leq M \int_{-\infty}^\infty |f(nx)| dx.$$
Question 3: Substituting $nx\mapsto t$ gives $$ \int_{-\infty}^\infty |f(nx)| dx = \frac{1}{n} \int_{-\infty}^\infty |f(t)| dt.$$
As $f$ is integrable, you can conclude. 
