0
$\begingroup$

Given that Let $C[0,1]$ be the Real vector space of all the continous Real valued function on $[0,1]$ let $T$ be the linear operator on this and defined by

$$(Tf)(x)=\int_0^1 \sin(x+y)f(y) \;dy\;\;,\;x \in [0,1]$$

find the Dimensions of Range space of $T$

Solution I tried-The given transformation is a operator So it will defined like $$T:C[0,1]\rightarrow C[0,1]$$ after that i have no idea how to solve ,Please provide me a hint

Thank you

Improve this question
$\endgroup$
3
$\begingroup$

$Tf(x)=\int_0^{1} [\sin x\cos y+\cos x \sin y ]f(y)dy=a\sin x+b \cos x$ where $a=\int_0^{1} \cos y f(y)dy$ and $b=\int_0^{1} \sin y f(y)dy$. Hence the range is contained in the two dimensional space spanned by $\sin x$ and $\cos x$. I will leave it to you to check that these two functions are actually in the range (or some linear combinations of these which are not scalar multiples of each other are in the range). Hence the range is 2-dimensional.

Improve this answer
$\endgroup$
  • $\begingroup$ That means i have to think about two $f(y)$ for which i get $a=1,b=0$ for 2nd one $a=0,b=1$ $\endgroup$ – honey kumar Jan 27 '20 at 5:46
  • 1
    $\begingroup$ Instead of that it is enough to find some linear combinations of $\sin x$ and $\cos x$ in the range which are not multiples of each other (so that they are linearly independent). For that it is enough to consider $f=1$ and $f(x)=x$. For $f=1$ it is easy to write down $Tf$. For $f(x)=x$ use integration by parts. @TheStudent $\endgroup$ – Kavi Rama Murthy Jan 27 '20 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.