# Show its a continuous linear functional (distribution)

I want to show that for $\phi\in C_0^{\infty}(\mathbb{R})$, the following map

$$\Lambda(\phi) = \int_{-a}^a\frac{\phi(x)-\phi(0)}{x}dx$$ for $a>0$, is a continuous linear functional. (i.e. a distribution). Linearity is clear.

So if $(K_j)$ is an increasing sequence of compact sets such that $\ \bigcup_j K_j = \mathbb{R}$, then the way to prove this, is to show that for all $j$, there exists positive number $N_j$ and consant $c_j$ such that $$|\Lambda(\phi)|\leq c_j\sup\left\{|\partial^{\alpha}\phi(x)| \ |x\in K_j, \alpha\leq N_j \right\}$$

for all $\phi\in C_0^{\infty}(K_j)$. With most integrals this is an easy estimation to make, but this $1/x$ thing makes things slightly problametic..

Can someone shed some light over this?

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Hint: as $\displaystyle\left|\frac{\phi(x)-\phi(0)}x\right|\leqslant \sup_{-a\leqslant t\leqslant a}|\phi'(t)|,$ we have $\displaystyle|\Lambda(\phi)|\leqslant 2a\cdot \sup_{-a\leqslant t\leqslant a}|\phi'(t)|$.
Sorry im little confused. If for $K_j$ this condition holds for $N_j=1$, shouldnt it also hold for $N_j=0$?. The inequality $|\Lambda(\phi)|\leq c_j \sup\left\{|\phi'(x)| \ |x\in K_j\right\}$ might hold, but then $c_j \sup \left\{|\phi(x)| \ |x\in K_j\right\}$ must also hold right for it to be continuous? I don't see this happening at all... – DinkyDoe Feb 22 '13 at 19:02