Distributions
*Definition
Let ${v_k} $ be a sequence of elements of $D(Omega)$ space and $ v in D(Omega)$.
We say that ${v_k}$ converges to $v in D(Omega)$ when the following conditions are verified:
a) It exists a compact $K sub Omega $ which includes the supports of all ${v_k}$.
b) For each operator $ D $ of partial differentiation (of whatevere order $>=0$) we have :
$lim_(k to oo) Dv_k(x) =Dv(x) $ uniformly in $Omega$.
That is to say in better words:
A sequence converges ${v_k} to v $ ,iff there is a compact set $K$ such that all $ v_k $ are supported in $K$ and every derivative $D^alpha v_k $ converges uniformly to $D^alpha v $ in $K$ .
*Exercise
Let be $phi in D(RR) $ not identically null and let put :
$u_k(t)=2^(-k) phi(t) ; ******* v_k(t)=2^(-k) phi(k^5*t) ;*********** w_k(t)=2^(-k)phi(t/k) $
$x_k(t)=2^(-k)phi(2^k*t) ; ******* y_k(t)=(1/k)phi(t-k) ; *********** z_k(t)=2^(-k)phi(t-k) $.
Determine if the six sequences converge or not to the null function in $D(RR)$ .
Let ${v_k} $ be a sequence of elements of $D(Omega)$ space and $ v in D(Omega)$.
We say that ${v_k}$ converges to $v in D(Omega)$ when the following conditions are verified:
a) It exists a compact $K sub Omega $ which includes the supports of all ${v_k}$.
b) For each operator $ D $ of partial differentiation (of whatevere order $>=0$) we have :
$lim_(k to oo) Dv_k(x) =Dv(x) $ uniformly in $Omega$.
That is to say in better words:
A sequence converges ${v_k} to v $ ,iff there is a compact set $K$ such that all $ v_k $ are supported in $K$ and every derivative $D^alpha v_k $ converges uniformly to $D^alpha v $ in $K$ .
*Exercise
Let be $phi in D(RR) $ not identically null and let put :
$u_k(t)=2^(-k) phi(t) ; ******* v_k(t)=2^(-k) phi(k^5*t) ;*********** w_k(t)=2^(-k)phi(t/k) $
$x_k(t)=2^(-k)phi(2^k*t) ; ******* y_k(t)=(1/k)phi(t-k) ; *********** z_k(t)=2^(-k)phi(t-k) $.
Determine if the six sequences converge or not to the null function in $D(RR)$ .
Risposte
Yes indeed , that's the proper solution . 
Maybe I should spend some more words to explain this exercise...
Of course, the sequences [1] and [2] verify both the (a) and (b) conditions.
The sequence [4] verifies only (a), but not (b).
The sequences [3], [5] and [6] instead verify only (b), but not (a).
Of course, the sequences [1] and [2] verify both the (a) and (b) conditions.
The sequence [4] verifies only (a), but not (b).
The sequences [3], [5] and [6] instead verify only (b), but not (a).
That's correct.
I'll try...
1- yes
2- yes
3- no
4- no
5- no
6- no
1- yes
2- yes
3- no
4- no
5- no
6- no
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