Once you get away from vector spaces of finite dimension, it is no longer true that every linear map is automatically continuous. For example, let Coo {f €F(N, R) f(n) = 0 for all sufficiently large n}, where F(N, R) is the vector space of all func tions from N to R with the usual pointwise operations. Another way of expressing the condition for f to belong to Coo is that {n: f(n) #0} is finite. Then define L: Coo R by L(f) = Enf(n). Notice that the sum on the right hand side is in fact a finite sum. Now (a) Check that L is linear.

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Once you get away from vector spaces of finite dimension, it is no longer true that
every linear map is automatically continuous. For example, let Coo {f E F(N, R) :
f(n) = 0 for all sufficiently large n}, where F(N, R) is the vector space of all func-
tions from N to R with the usual pointwise operations. Another way of expressing
the condition for f to belong to Coo is that {n : f(n) # 0} is finite.
Then define L: C00R by L(f) = Enf(n).
Notice that the sum on the right hand side is in fact a finite sum. Now
(a) Check that L is linear.
(b) Show that {|L(f)| : ||F|| < 1} is not bounded above. Then deduce that L is not
continuous, from the following theorem:
For each continuous linear map L:V W the set {I|L(x)|lw: x|lv <1}
is bounded above.
1 n=r
(Hint: consider the functions e,, defined by e, (n) =
and define a
%3D
0 otherwise
norm on e, by ||e,|| := sup{e,(n)|: n < 1} = le,(1)|. Show that for all r E N,
|le,|| < 1. On the other hand {|L(e,)|:r€ N} is not bounded above.]
Transcribed Image Text:Once you get away from vector spaces of finite dimension, it is no longer true that every linear map is automatically continuous. For example, let Coo {f E F(N, R) : f(n) = 0 for all sufficiently large n}, where F(N, R) is the vector space of all func- tions from N to R with the usual pointwise operations. Another way of expressing the condition for f to belong to Coo is that {n : f(n) # 0} is finite. Then define L: C00R by L(f) = Enf(n). Notice that the sum on the right hand side is in fact a finite sum. Now (a) Check that L is linear. (b) Show that {|L(f)| : ||F|| < 1} is not bounded above. Then deduce that L is not continuous, from the following theorem: For each continuous linear map L:V W the set {I|L(x)|lw: x|lv <1} is bounded above. 1 n=r (Hint: consider the functions e,, defined by e, (n) = and define a %3D 0 otherwise norm on e, by ||e,|| := sup{e,(n)|: n < 1} = le,(1)|. Show that for all r E N, |le,|| < 1. On the other hand {|L(e,)|:r€ N} is not bounded above.]
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