\(\blacksquare \) Definition of pointwise convergence: \(f_{n}\left ( x\right ) \) converges pointwise to \(f_{\ast }\left ( x\right ) \) if for each \(\varepsilon >0\) there exist integer \(N\left ( \varepsilon ,x\right ) \) such that \(\left \vert f_{n}\left ( x\right ) -f_{\ast }\left ( x\right ) \right \vert <\varepsilon \) for all \(n\geq N\).
\(\blacksquare \) Definition of uniform convergence: \(f_{n}\left ( x\right ) \) converges uniformly to \(f_{\ast }\left ( x\right ) \) if for each \(\varepsilon >0\) there exist integer \(N\left ( \varepsilon \right ) \) such that \(\left \vert f_{n}\left ( x\right ) -f_{\ast }\left ( x\right ) \right \vert <\varepsilon \) for all \(n\geq N\).
\(\blacksquare \) Another way to find uniform convergence, first find pointwise convergence of \(f_{n}\left ( x\right ) \). Say it converges to \(f_{\ast }\left ( x\right ) \). Now show that
goes to zero as \(n\rightarrow \infty \). To find \(\sup \left ( f_{n}-f_{\ast }\right ) \) might need to find the maximum of \(f_{n}-f_{\ast }\). i.e. differentiate this, set to zero, find \(x\) where it is Max, then evaluate \(f_{n}\left ( x\right ) -f_{\ast }\left ( x\right ) \) at this maximum. This gives the \(\sup \). Then see if this goes to zero as \(n\rightarrow \infty \)
\(\blacksquare \) If sequence of functions \(f_{n}\) converges uniformly to \(f_{\ast }\), then \(f_{\ast }\) must be continuous. So this gives a quick check if uniform convergence exist. First find the pointwise convergence \(f_{\ast }\left ( x\right ) \) and check if this is continuous or not. If not, then no need to check for uniform convergence, it does not exist. But if \(f_{\ast }\left ( x\right ) \) is continuous function, we still need to check because it is possible there is no uniform convergence.