let $ H= L^2[0,1]$ and $ C^1 $ be the set of all continuouse functions on $ [0,1] $ that have continuouse derivative.Let $ t \in [0,1] $ and define $ L: C^1 \longrightarrow F $ by $ L (h)= h'(t) $. show that there is no bounded linear functional on $ H $ that agrees with $ L $ on $ H $ .thanks
Asked
Active
Viewed 388 times
2
-
What is the space $F$? – gerw May 20 '13 at 18:01
-
Hi!, what sequence of functions {fn}∈C1{fn}∈C1, we can use?? I think, fn(x)=sin(nx)/nfn(x)=sin(nx)/n. Regards! – MathUser Oct 12 '16 at 01:48
1 Answers
4
Hint: Try to find a sequence of functions $\{f_n\}$ in $C^1$, such that $f_n$ is bounded in $H$ (i.e., the function values should not be large), but $f_n'$ unbounded in $F$ (i.e., the derivative is large).
gerw
- 31,359
-
Hi, what sequence of functions ${f_n}\in\mathcal{C}^1$, we can use?? I think, $f_n(x)=sin(nx)/n$. Regards! – MathUser Oct 12 '16 at 01:47
-
This might work but will be tedious to check. I think it is easier to work with piecewise quadratic polynomials. – gerw Oct 12 '16 at 09:29