Answer to Question #322804 in Functional Analysis for Deepu

Question #322804

For an example of a discontinuous linear functional on a normed linear space


1
Expert's answer
2022-04-04T15:53:09-0400

The  linear  space  is  LC1[0,1]  of  continuous  functions  on  [0,1]  withx=01x(t)dtThe  functional  isf(x)=x(1)Thenf(αx+βy)=(αx+βy)(1)=αx(1)+βy(1)=αf(x)+βf(y)thus  f  is  linear.f  is  not  continuous,since  for  xn(t)=tnxn=01tndt=1n+10,nwhich  means  xn0,nin  LC1[0,1]And  we  havef(xn)=1n=1,f(0)=0thusf(xn)f(0),nwhich  means  f  is  discontinuousThe\,\,linear\,\,space\,\,is\,\,LC_1\left[ 0,1 \right] \,\,of\,\,continuous\,\,functions\,\,on\,\,\left[ 0,1 \right] \,\,with\\\left\| x \right\| =\int_0^1{\left| x\left( t \right) \right|dt}\\The\,\,functional\,\,is\\f\left( x \right) =x\left( 1 \right) \\Then\\f\left( \alpha x+\beta y \right) =\left( \alpha x+\beta y \right) \left( 1 \right) =\alpha x\left( 1 \right) +\beta y\left( 1 \right) =\alpha f\left( x \right) +\beta f\left( y \right) \\thus\,\,f\,\,is\,\,linear.\\f\,\,is\,\,not\,\,continuous, \sin ce\,\,for\,\,\\x_n\left( t \right) =t^n\\\left\| x_n \right\| =\int_0^1{\left| t^n \right|dt}=\frac{1}{n+1}\rightarrow 0,n\rightarrow \infty \\which\,\,means\,\,x_n\rightarrow 0,n\rightarrow \infty in\,\,LC_1\left[ 0,1 \right] \\And\,\,we\,\,have\\f\left( x_n \right) =1^n=1,f\left( 0 \right) =0\\thus\\f\left( x_n \right) \nrightarrow f\left( 0 \right) ,n\rightarrow \infty \\which\,\,means\,\,f\,\,is\,\,discontinuous


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