gn = gn ⁢1 gn ⁢2=1+1n2−1n gn gn 1 gn 2 1 1 n 2 1 n First we find the following limits for our two gn gn's: limit   n → ∞ gn ⁢1=1 n gn 1 1 limit   n → ∞ gn ⁢2=2 n gn 2 2 Therefore we have the following, limit   n → ∞ gn =g n gn g pointwise, where g=12 g 1 2 .

∀ t ,t∈R: gn ⁢t=tn t t gn t t n As done above, we first want to examine the limit limit   n → ∞ gn ⁢t0=limit   n → ∞ t0n=0 n gn t0 n t0 n 0 where t0∈R t0 . Thus limit   n → ∞ gn=g n gn g pointwise where g⁢t=0 g t 0 for all t∈R t .

g n =1+1n2−1n g n 1 1 n 2 1 n Let g=12 g 1 2

g n ⁢t={tn  if  0≤t≤10  otherwise   g n t t n 0 t 1 0 Let g⁢t=0 g t 0 for all tt.

We are given the following function: g n ⁢t={n  if  0<t<1n0  otherwise   g n t n 0 t 1 n 0 Then limit   n → ∞ g n ⁢t=0 n g n t 0 This means that, g n ⁢t→g⁢t g n t g t pointwise where for all tt g⁢t=0 g t 0 .

Now,

We are given the following function: g n ⁢t={1  if  0<t<1n0  otherwise   if n is even g n t 1 0 t 1 n 0 if n is even g n ⁢t={-1  if  0<t<1n0  otherwise   if n is odd g n t -1 0 t 1 n 0 if n is odd Then, ∥ g n −g∥=∫01n1d t =1n→0 g n g t 1 n 0 1 1 n 0 where g⁢t=0 g t 0 for all tt. Therefore, g n →g in norm g n g in norm However, at t=0 t 0 , g n ⁢t g n t oscillates between -1 and 1, and so it does not converge. Thus, g n ⁢t g n t does not converge pointwise.