Properties of bounded linear operators between two normed spaces












0















Let $T$ be a linear operator between two normed spaces $X$ and $Y$. Show that





  • a) If $T$ is not bounded then for $varepsilon>0$ , $sup Vert TxVert = infty$ where $Vert x Vert < varepsilon$.

  • b) If ${T(x_{n})}$ is bounded for any sequence ${x_{n}}$ such that $x_nto 0$, then $T$ is bounded.


Please help me if you have a good idea for my question. I do not understand this concept. Thank you.






functional-analysis linear-transformations normed-spaces







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  • @mathworker21 it's not nice to say that. The question is perfectly legit.
    – John_Wick
    Nov 24 at 14:15










  • @John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
    – mathworker21
    Nov 24 at 14:16










  • Never mind. I apologize if I have offended you.
    – John_Wick
    Nov 24 at 14:17
















0















Let $T$ be a linear operator between two normed spaces $X$ and $Y$. Show that





  • a) If $T$ is not bounded then for $varepsilon>0$ , $sup Vert TxVert = infty$ where $Vert x Vert < varepsilon$.

  • b) If ${T(x_{n})}$ is bounded for any sequence ${x_{n}}$ such that $x_nto 0$, then $T$ is bounded.


Please help me if you have a good idea for my question. I do not understand this concept. Thank you.






functional-analysis linear-transformations normed-spaces







share|cite|improve this question
























  • @mathworker21 it's not nice to say that. The question is perfectly legit.
    – John_Wick
    Nov 24 at 14:15










  • @John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
    – mathworker21
    Nov 24 at 14:16










  • Never mind. I apologize if I have offended you.
    – John_Wick
    Nov 24 at 14:17














0












0








0








Let $T$ be a linear operator between two normed spaces $X$ and $Y$. Show that





  • a) If $T$ is not bounded then for $varepsilon>0$ , $sup Vert TxVert = infty$ where $Vert x Vert < varepsilon$.

  • b) If ${T(x_{n})}$ is bounded for any sequence ${x_{n}}$ such that $x_nto 0$, then $T$ is bounded.


Please help me if you have a good idea for my question. I do not understand this concept. Thank you.






functional-analysis linear-transformations normed-spaces







share|cite|improve this question
















Let $T$ be a linear operator between two normed spaces $X$ and $Y$. Show that





  • a) If $T$ is not bounded then for $varepsilon>0$ , $sup Vert TxVert = infty$ where $Vert x Vert < varepsilon$.

  • b) If ${T(x_{n})}$ is bounded for any sequence ${x_{n}}$ such that $x_nto 0$, then $T$ is bounded.


Please help me if you have a good idea for my question. I do not understand this concept. Thank you.






functional-analysis linear-transformations normed-spaces




functional-analysis linear-transformations normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 21:46









user587192

1,488112




1,488112










asked Nov 24 at 13:17









ezgialtinbasak

15




15












  • @mathworker21 it's not nice to say that. The question is perfectly legit.
    – John_Wick
    Nov 24 at 14:15










  • @John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
    – mathworker21
    Nov 24 at 14:16










  • Never mind. I apologize if I have offended you.
    – John_Wick
    Nov 24 at 14:17


















  • @mathworker21 it's not nice to say that. The question is perfectly legit.
    – John_Wick
    Nov 24 at 14:15










  • @John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
    – mathworker21
    Nov 24 at 14:16










  • Never mind. I apologize if I have offended you.
    – John_Wick
    Nov 24 at 14:17
















@mathworker21 it's not nice to say that. The question is perfectly legit.
– John_Wick
Nov 24 at 14:15




@mathworker21 it's not nice to say that. The question is perfectly legit.
– John_Wick
Nov 24 at 14:15












@John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
– mathworker21
Nov 24 at 14:16




@John_Wick i think you could have guessed he and I had a dialogue and then he deleted his comments
– mathworker21
Nov 24 at 14:16












Never mind. I apologize if I have offended you.
– John_Wick
Nov 24 at 14:17




Never mind. I apologize if I have offended you.
– John_Wick
Nov 24 at 14:17










1 Answer
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oldest

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(a) If $sup_{|x|<epsilon} |Tx|=M<infty $ then $sup_{xin X,|x|>0}frac{|Tx|}{|x|}=sup_{xin X,|x|>0}|T(epsilon x/|x|)|/epsilonleq sup_{|x|<epsilon} |Tx|/epsilon=M/epsilon<infty. $ Hence $sup_{|x|<epsilon} |Tx|<infty$ implies $T$ is bounded.



(b) Let $T$ is unbounded. So, by (a) for any $ninmathbb{N}$, $sup_{|x|<1/n}=infty$. So there exists an $x_n$ with $|x_n|<1/n$ such that $|T(x_n)|geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $|T(x_n)|rightarrow infty.$






share|cite|improve this answer























  • you could be dividing by 0 at one step in part a
    – mathworker21
    Nov 24 at 13:49












  • Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
    – John_Wick
    Nov 24 at 14:13










  • Thank you John really.
    – ezgialtinbasak
    Nov 24 at 14:22











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1 Answer
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(a) If $sup_{|x|<epsilon} |Tx|=M<infty $ then $sup_{xin X,|x|>0}frac{|Tx|}{|x|}=sup_{xin X,|x|>0}|T(epsilon x/|x|)|/epsilonleq sup_{|x|<epsilon} |Tx|/epsilon=M/epsilon<infty. $ Hence $sup_{|x|<epsilon} |Tx|<infty$ implies $T$ is bounded.



(b) Let $T$ is unbounded. So, by (a) for any $ninmathbb{N}$, $sup_{|x|<1/n}=infty$. So there exists an $x_n$ with $|x_n|<1/n$ such that $|T(x_n)|geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $|T(x_n)|rightarrow infty.$






share|cite|improve this answer























  • you could be dividing by 0 at one step in part a
    – mathworker21
    Nov 24 at 13:49












  • Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
    – John_Wick
    Nov 24 at 14:13










  • Thank you John really.
    – ezgialtinbasak
    Nov 24 at 14:22
















0














(a) If $sup_{|x|<epsilon} |Tx|=M<infty $ then $sup_{xin X,|x|>0}frac{|Tx|}{|x|}=sup_{xin X,|x|>0}|T(epsilon x/|x|)|/epsilonleq sup_{|x|<epsilon} |Tx|/epsilon=M/epsilon<infty. $ Hence $sup_{|x|<epsilon} |Tx|<infty$ implies $T$ is bounded.



(b) Let $T$ is unbounded. So, by (a) for any $ninmathbb{N}$, $sup_{|x|<1/n}=infty$. So there exists an $x_n$ with $|x_n|<1/n$ such that $|T(x_n)|geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $|T(x_n)|rightarrow infty.$






share|cite|improve this answer























  • you could be dividing by 0 at one step in part a
    – mathworker21
    Nov 24 at 13:49












  • Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
    – John_Wick
    Nov 24 at 14:13










  • Thank you John really.
    – ezgialtinbasak
    Nov 24 at 14:22














0












0








0






(a) If $sup_{|x|<epsilon} |Tx|=M<infty $ then $sup_{xin X,|x|>0}frac{|Tx|}{|x|}=sup_{xin X,|x|>0}|T(epsilon x/|x|)|/epsilonleq sup_{|x|<epsilon} |Tx|/epsilon=M/epsilon<infty. $ Hence $sup_{|x|<epsilon} |Tx|<infty$ implies $T$ is bounded.



(b) Let $T$ is unbounded. So, by (a) for any $ninmathbb{N}$, $sup_{|x|<1/n}=infty$. So there exists an $x_n$ with $|x_n|<1/n$ such that $|T(x_n)|geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $|T(x_n)|rightarrow infty.$






share|cite|improve this answer














(a) If $sup_{|x|<epsilon} |Tx|=M<infty $ then $sup_{xin X,|x|>0}frac{|Tx|}{|x|}=sup_{xin X,|x|>0}|T(epsilon x/|x|)|/epsilonleq sup_{|x|<epsilon} |Tx|/epsilon=M/epsilon<infty. $ Hence $sup_{|x|<epsilon} |Tx|<infty$ implies $T$ is bounded.



(b) Let $T$ is unbounded. So, by (a) for any $ninmathbb{N}$, $sup_{|x|<1/n}=infty$. So there exists an $x_n$ with $|x_n|<1/n$ such that $|T(x_n)|geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $|T(x_n)|rightarrow infty.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 14:14

























answered Nov 24 at 13:44









John_Wick

1,314111




1,314111












  • you could be dividing by 0 at one step in part a
    – mathworker21
    Nov 24 at 13:49












  • Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
    – John_Wick
    Nov 24 at 14:13










  • Thank you John really.
    – ezgialtinbasak
    Nov 24 at 14:22


















  • you could be dividing by 0 at one step in part a
    – mathworker21
    Nov 24 at 13:49












  • Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
    – John_Wick
    Nov 24 at 14:13










  • Thank you John really.
    – ezgialtinbasak
    Nov 24 at 14:22
















you could be dividing by 0 at one step in part a
– mathworker21
Nov 24 at 13:49






you could be dividing by 0 at one step in part a
– mathworker21
Nov 24 at 13:49














Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
– John_Wick
Nov 24 at 14:13




Oh, for that take $x$ such that $|x|>0.$ I am going to edit my answer.
– John_Wick
Nov 24 at 14:13












Thank you John really.
– ezgialtinbasak
Nov 24 at 14:22




Thank you John really.
– ezgialtinbasak
Nov 24 at 14:22


















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