## Some questions

Forum for the GRE subject test in mathematics.
Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

### Some questions

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)

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
I'll try to answer Q.1 only. From the description it seems that the f(x) will form a triangle with the vertices located at (0,0), (2, 0) and (k, 1). The area will be a constant [ A = 1/2 * base * height = 1/2 * 2 * 1 = 1].

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]

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Good job amateur,
this is a problem of previous GRE maths test . I don't have the answer key . But the idea is your idea

let kill Q.2, I think we can find some counter examples for this problem but I have not found out yet

blp
Posts: 17
Joined: Sun Oct 19, 2008 5:24 am
There's a pretty typical counterexample for the second problem, though writing down the specific equation is a bit harder.

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.