Is the functional $fmapstoint_0^1 f(x)x^{-1/2}dx$ continuous in $L^2[0,1]$?












2














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







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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
















2














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







share|cite|improve this question
























  • 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














2












2








2







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 9:56









Did

246k23220454




246k23220454










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


















  • 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















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