I'm trying to prove what follows: if $f \in L^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$, with $\hat{f}(x) = \int_{\mathbb{R}} f(t) e^{-ixt} dt$, and $\phi_n(x) = \frac{n}{\pi}\frac{1}{1+n^2x^2}$ for $n \in \mathbb{N}$ and $x \in \mathbb{R}$, does the convolution product $(f*\phi_n)(x) \to f(x)$ for every $x \in \mathbb{R}$? Is there uniform convergence?
The point-wise convergence is quite easy as it suffices to neatly apply the inversion theorem for the Fourier transform. I'm having problems with proving or disproving the last part - uniform convergence.
I know that $\phi_n \in C_0(\mathbb{R})$ since $\phi_n,\hat{\phi_n} \in L^1(\mathbb{R})$ and thus $\phi_n$ is uniformly continuous for each $n \in \mathbb{N}$. Going this way hasn't led me anywhere though.
Would you give me any hint on how I should approach this? Please accept my excuses for how trivial this might be, but I haven't got anywhere.
The convegergence is uniform. The inversion theorem shows that $f$ is bounded and uniformly continuous. [In fact it is continuous and $f(x) \to 0$ as $|x| \to \infty$]. Hence $$|f*\phi_n)x)-f(x)|=|\int f(x-y)\phi_n(y)dy-f(x)|$$ $$=|\int [f(x-y)-f(x)]\phi_n(y)dy| \leq \int |f(x-y)-f(x)|\phi_n(y)dy$$
Let $\epsilon >0$ and choose $\delta $ such that $|f(x-y)-f(y)| <\epsilon$ for $|y| <\delta$. Now split the integral over $y$ integral over $|y| <\delta$ and $|y| >\delta$. Use the boundedness of $f$ in the second part. Can you finish?