Convolution with standard mollifier Let $\Omega \subset \mathbb{R}$ open and $f \in L^p(\Omega).$
Now, we define $$\eta(x):=\chi_{[-1,1]}(x) e^{\frac{-1}{1-x^2}}.$$
Then we define $$\eta_h(x):=\frac{1}{h} \eta\left( \frac{x}{h}\right).$$ This means that $\mathrm{supp}(\eta_h) \subset [-h,h].$
Now, we clearly have $f * \eta_h \in L^p(\Omega).$ Does anybody know if we have $$||f * \eta_h ||_p \le ||f||$$ and $$||f - f* \eta_h || \rightarrow 0$$ for $h \rightarrow \infty, $ too?

There must be a typo; you mean $\eta_h(x)=\frac1h\eta(\frac xh)$, right? Assuming so: I have no idea why you're interested in that particular function, nor what its integral is. Say $$\alpha=\int\eta.$$
Young's inequality says in part that $||f*g||_p\le||f||_p||g||_1$; since $\eta\ge0$ this shows that $$||f*\eta_h||_p\le\alpha||f||_p.$$
And standard results/techniques show easily that $$||\alpha f-f*\eta_h||_p\to0.$$
