- #1
BaitiTamam
- 4
- 0
Let ρ(x) be a continuous function on ℝ, which evaluates ρ(x)=0 when |x|≥1 and that meets following.
∫[-1,1]ρ(x)dx=1
And let ψ(x) be a continuous function on interval [-1,1], prove
lim[n→∞] n∫[-1,1]ρ(nx)ψ(x)dx = ψ(0).
is denoted.
This is NOT a homework but a past exam problem of a college that has no answer.
I've been thinking for a whole day now.
What I have found so far:
First I defined sequencing functions as following:
F_n(x)=∫[-1,x]ρ(ny)ψ(y)dy
G_n(x)=∫[x,1]ρ(ny)ψ(y)dy
Now given a properly large natural number of n_0 which completes n_0≥1/y (where y≠0),
for all n∈N of n≥n_0, ρ(ny) evaluates 0. In this sense, taken a small positive number ε>0 instead of y, we only have to concern that of a small closed interval of [-ε,ε] as to nF_n(-ε) and nG_n(ε) gives 0 (for all n≥n_0≥1/ε).
So for all n same above,
n∫[-1,1]ρ(nx)ψ(x)dx = n∫[-ε,ε]ρ(nx)ψ(x)dx
And I have a stuck.
Please help ;)
∫[-1,1]ρ(x)dx=1
And let ψ(x) be a continuous function on interval [-1,1], prove
lim[n→∞] n∫[-1,1]ρ(nx)ψ(x)dx = ψ(0).
is denoted.
This is NOT a homework but a past exam problem of a college that has no answer.
I've been thinking for a whole day now.
What I have found so far:
First I defined sequencing functions as following:
F_n(x)=∫[-1,x]ρ(ny)ψ(y)dy
G_n(x)=∫[x,1]ρ(ny)ψ(y)dy
Now given a properly large natural number of n_0 which completes n_0≥1/y (where y≠0),
for all n∈N of n≥n_0, ρ(ny) evaluates 0. In this sense, taken a small positive number ε>0 instead of y, we only have to concern that of a small closed interval of [-ε,ε] as to nF_n(-ε) and nG_n(ε) gives 0 (for all n≥n_0≥1/ε).
So for all n same above,
n∫[-1,1]ρ(nx)ψ(x)dx = n∫[-ε,ε]ρ(nx)ψ(x)dx
And I have a stuck.
Please help ;)