The function $u(\sigma,t)$ is defined by $u(\sigma,t)=\frac{1}{\pi}\int_{-\infty}^{\infty}{f(x)\frac{\sigma dx}{\sigma^2+(x-t)^2}}=\frac{1}{\pi}\int_{-\infty}^{\infty}{f(t+\sigma u)\frac{du}{1+u^2}}$, where $\sigma u=x-t$, and f(t) is uniformly continuous and bounded. I'm trying to prove that $u(\sigma,t)$ converges to $f(t)$ as $\sigma$ approaches to zero, uniformly in $t$. I write: $$|u(\sigma_m,t)-u(\sigma_n,t)|=\frac{1}{\pi}\left|\int_{-\infty}^{\infty}{\frac{du}{1+u^2}[f(t+\sigma_mu)-f(t+\sigma_nu)]}\right|$$ The integral converges so we can find $N>0$ large enough to ensure that both $\frac{1}{\pi}|\int_{-\infty}^{-N}|$ and $\frac{1}{\pi}|\int_N^{\infty}|$ are lower than $\frac{\varepsilon}{3}$, and thus: $$|u(\sigma_m,t)-u(\sigma_n,t)|<\frac{1}{\pi}\left|\int_{-N}^{N}{\frac{du}{1+u^2}[f(t+\sigma_mu)-f(t+\sigma_nu)]}\right|+\frac{2\varepsilon}{3}$$ now, since that $f$ is uniformly continuous, we can find $\delta(\varepsilon/3)$ so that if $|t_1-t_2|<\delta(\varepsilon/3)$, then $|f(t_1)-f(t_2)|<\varepsilon/3$. If $\sigma_m,\sigma_n<\frac{\delta(\varepsilon/3)}{N}$ then the next inequality holds: $$\frac{1}{\pi}\left|\int_{-N}^{N}{\frac{du}{1+u^2}[f(t+\sigma_mu)-f(t+\sigma_nu)]}\right|<\frac{1}{\pi}\int_{-N}^{N}{\frac{du}{1+u^2}|f(t+\sigma_mu)-f(t+\sigma_nu)|}<\frac{\varepsilon}{3\pi}\left[arctan(u)\right]^{N}_{-N}<\frac{\varepsilon}{3\pi}\pi=\frac{\varepsilon}{3}$$ and to conclude: $$|u(\sigma_m,t)-u(\sigma_n,t)|<\varepsilon$$ And thus by Cauchy's criterion, the sequence converges uniformly. Now I'm not sure- does that proof hold?
Edit 1: Now I see that even if the proof holds, that I only showed that it converges uniformly to some function, not particularly to $f(t)$. Suppose that the proof above is legal, I'm again not sure whether is it suffice to show that $u(\sigma,t)$ converges point-wise to $f(t)$ this way: $$|\frac{1}{\pi}\int_{-\infty}^{\infty}{f(t+\sigma u)\frac{du}{1+u^2}}|<\frac{1}{\pi}\int_{-N}^{N}{|f(t+\sigma u)|\frac{du}{1+u^2}}+\frac{2}{3\varepsilon}<\frac{|f(t)|+(\varepsilon/3)}{\pi}\int_{-N}^N{\frac{du}{1+u^2}}<\varepsilon$$