Is the functional $fmapstoint_0^1 f(x)x^{-1/2}dx$ continuous in $L^2[0,1]$?
I was looking for examples of linear functionals which are continuous in $L^2[a,b]$ but not $L^1[a,b]$. My first thought was to try $F(f)=int_0^1f(x)x^{-1/2}dx$, but I had trouble showing it was continuous in $L^2[0,1]$. However, it was easy to show that $fmapsto int_0^1f(x)x^{-1/3}dx$ was an example by using H$ddot{text{o}}$lder's inequality (or more specifically, the Cauchy-Schwarz inequality in this case). After some thought, I concluded that the linear functional $fmapstoint_0^1f(x)x^{-1/2+epsilon}dx$ is continuous and $fmapstoint_0^1f(x)x^{-1/2-epsilon}dx$ is discontinuous in $L^2[0,1]$ for any $epsilonin(0,1/2)$. However, I'm still a little stuck on determining whether $F(f)=int_0^1f(x)x^{-1/2}dx$ is continuous. My instincts tell me it should be continuous, but H$ddot{text{o}}$lder's inequality doesn't work as nicely as it did when the exponent on $x$ is greater than $-1/2$.
functional-analysis lp-spaces
edited Nov 26 at 9:56
Did
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
add a comment |
I was looking for examples of linear functionals which are continuous in $L^2[a,b]$ but not $L^1[a,b]$. My first thought was to try $F(f)=int_0^1f(x)x^{-1/2}dx$, but I had trouble showing it was continuous in $L^2[0,1]$. However, it was easy to show that $fmapsto int_0^1f(x)x^{-1/3}dx$ was an example by using H$ddot{text{o}}$lder's inequality (or more specifically, the Cauchy-Schwarz inequality in this case). After some thought, I concluded that the linear functional $fmapstoint_0^1f(x)x^{-1/2+epsilon}dx$ is continuous and $fmapstoint_0^1f(x)x^{-1/2-epsilon}dx$ is discontinuous in $L^2[0,1]$ for any $epsilonin(0,1/2)$. However, I'm still a little stuck on determining whether $F(f)=int_0^1f(x)x^{-1/2}dx$ is continuous. My instincts tell me it should be continuous, but H$ddot{text{o}}$lder's inequality doesn't work as nicely as it did when the exponent on $x$ is greater than $-1/2$.
functional-analysis lp-spaces
edited Nov 26 at 9:56
Did
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
1
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59
add a comment |
I was looking for examples of linear functionals which are continuous in $L^2[a,b]$ but not $L^1[a,b]$. My first thought was to try $F(f)=int_0^1f(x)x^{-1/2}dx$, but I had trouble showing it was continuous in $L^2[0,1]$. However, it was easy to show that $fmapsto int_0^1f(x)x^{-1/3}dx$ was an example by using H$ddot{text{o}}$lder's inequality (or more specifically, the Cauchy-Schwarz inequality in this case). After some thought, I concluded that the linear functional $fmapstoint_0^1f(x)x^{-1/2+epsilon}dx$ is continuous and $fmapstoint_0^1f(x)x^{-1/2-epsilon}dx$ is discontinuous in $L^2[0,1]$ for any $epsilonin(0,1/2)$. However, I'm still a little stuck on determining whether $F(f)=int_0^1f(x)x^{-1/2}dx$ is continuous. My instincts tell me it should be continuous, but H$ddot{text{o}}$lder's inequality doesn't work as nicely as it did when the exponent on $x$ is greater than $-1/2$.
functional-analysis lp-spaces
edited Nov 26 at 9:56
Did
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
I was looking for examples of linear functionals which are continuous in $L^2[a,b]$ but not $L^1[a,b]$. My first thought was to try $F(f)=int_0^1f(x)x^{-1/2}dx$, but I had trouble showing it was continuous in $L^2[0,1]$. However, it was easy to show that $fmapsto int_0^1f(x)x^{-1/3}dx$ was an example by using H$ddot{text{o}}$lder's inequality (or more specifically, the Cauchy-Schwarz inequality in this case). After some thought, I concluded that the linear functional $fmapstoint_0^1f(x)x^{-1/2+epsilon}dx$ is continuous and $fmapstoint_0^1f(x)x^{-1/2-epsilon}dx$ is discontinuous in $L^2[0,1]$ for any $epsilonin(0,1/2)$. However, I'm still a little stuck on determining whether $F(f)=int_0^1f(x)x^{-1/2}dx$ is continuous. My instincts tell me it should be continuous, but H$ddot{text{o}}$lder's inequality doesn't work as nicely as it did when the exponent on $x$ is greater than $-1/2$.
functional-analysis lp-spaces
functional-analysis lp-spaces
edited Nov 26 at 9:56
Did
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
edited Nov 26 at 9:56
Did
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
edited Nov 26 at 9:56
Did
246k23220454
edited Nov 26 at 9:56
Did
246k23220454
edited Nov 26 at 9:56
Did
246k23220454
246k23220454
asked Nov 26 at 9:47
Julian Benali
24713
asked Nov 26 at 9:47
Julian Benali
24713
asked Nov 26 at 9:47
Julian Benali
24713
24713
It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
1
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59
add a comment |
It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
1
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59
It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
1
1
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59
add a comment |
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It is not continuous, since $x^{-1/2}notin L^2((0,1))$.
– PhoemueX
Nov 26 at 9:53
1
For a specific counterexample, try $$f_n(x)=frac{mathbf 1_{nx>1}}{sqrt xcdotlog x}$$
– Did
Nov 26 at 9:55
@Did Thank you! That's perfect!
– Julian Benali
Nov 26 at 9:59