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
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]
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]
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.
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.