1. The function f is a linear function composed of two linear pieces and continuous functions on [0,2], f(0)=f(2)=0, and max f =1. What is the length of the graph of f?

2. f_n and f are continuous functions and f_n(x)-------->f(x) pointwise .

let :

F_n(x)=int(f_n(t)dt)

F(x)=int(f(t)dt)

then which of the following is/ are correct :

a. in_{0}^{x}(F_n(t)dt)------->in_{0}^{x}(F(t)dt)

b. F'_n(x)------->f(x)

c. in_{0}^{x}(f_n(t)dt)------->in_{0}^{x}(f(t)dt)

## Some questions

Regarding the length of f(x), it will be

Length = sqrt(1 + k^2) + sqrt(1 + (k - 2)^2)

Length_maximum = 1 + sqrt(5) [For k = 0, 2]

Length_minimum = 2*sqrt(2) [For k = 1]

So think about a function that is otherwise constantly zero, but it contains a thin very high smooth "spike" which has area one. We can assume that for f_n the spike is e.g. centered around (n-1)/n and has width 1/n^2 with height large enough for the integral over it to be 1 (draw a picture).

Now if we take any point x in R. If x>=1, then for all n f_n(x)=0 and for all x<1 the spike has gone "past" x for large enough n, so that there's a k>0 with f_n(x)=0 if n>k. This means that f_n->0 pointwise. But we have

int_0^1 f_n = 1 for all n

It means that F_n(x)=1 for all x>=1 while F(x)=0 for all x. What this means is that (a) and (c) are incorrect, but (b) is correct because F'_n=f_n which converges to f=0 pointwise. Thus assuming that the limit function is continuous doesn't make the problem behave any better, so we really need uniform convergence.