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Errors in showing this function is well-defined.
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Let $X$ be a normed space and let $Y$ be a dense linear subspace of $X$. Define $phi:X^*rightarrow Y^*$ by $phi(f)=f|_Y$. Then I want to show $phi$ is well-defined. And here is my attempt:
For all $fin X^*$, $f|_Y$ is a continuous linear functional on $Y$; that is, $f|_Yin Y^*$. If $f_1=f_2in X^*$, then $phi(f_1)=f_1|_Y=f_2|_Y=phi(f_2)$. Hence, $phi:X^*rightarrow Y^*$ is well -defined.
But my professor said my proof was wrong in logic. It seems ok to me. Can anyone help me? Thank you!
functional-analysis normed-spaces
asked Jul 20 at 21:58
Answer Lee
48838
add a comment |Â
up vote
1
down vote
favorite
Let $X$ be a normed space and let $Y$ be a dense linear subspace of $X$. Define $phi:X^*rightarrow Y^*$ by $phi(f)=f|_Y$. Then I want to show $phi$ is well-defined. And here is my attempt:
For all $fin X^*$, $f|_Y$ is a continuous linear functional on $Y$; that is, $f|_Yin Y^*$. If $f_1=f_2in X^*$, then $phi(f_1)=f_1|_Y=f_2|_Y=phi(f_2)$. Hence, $phi:X^*rightarrow Y^*$ is well -defined.
But my professor said my proof was wrong in logic. It seems ok to me. Can anyone help me? Thank you!
functional-analysis normed-spaces
asked Jul 20 at 21:58
Answer Lee
48838
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a normed space and let $Y$ be a dense linear subspace of $X$. Define $phi:X^*rightarrow Y^*$ by $phi(f)=f|_Y$. Then I want to show $phi$ is well-defined. And here is my attempt:
For all $fin X^*$, $f|_Y$ is a continuous linear functional on $Y$; that is, $f|_Yin Y^*$. If $f_1=f_2in X^*$, then $phi(f_1)=f_1|_Y=f_2|_Y=phi(f_2)$. Hence, $phi:X^*rightarrow Y^*$ is well -defined.
But my professor said my proof was wrong in logic. It seems ok to me. Can anyone help me? Thank you!
functional-analysis normed-spaces
asked Jul 20 at 21:58
Answer Lee
48838
Let $X$ be a normed space and let $Y$ be a dense linear subspace of $X$. Define $phi:X^*rightarrow Y^*$ by $phi(f)=f|_Y$. Then I want to show $phi$ is well-defined. And here is my attempt:
For all $fin X^*$, $f|_Y$ is a continuous linear functional on $Y$; that is, $f|_Yin Y^*$. If $f_1=f_2in X^*$, then $phi(f_1)=f_1|_Y=f_2|_Y=phi(f_2)$. Hence, $phi:X^*rightarrow Y^*$ is well -defined.
But my professor said my proof was wrong in logic. It seems ok to me. Can anyone help me? Thank you!
functional-analysis normed-spaces
asked Jul 20 at 21:58
Answer Lee
48838
asked Jul 20 at 21:58
Answer Lee
48838
asked Jul 20 at 21:58
Answer Lee
48838
asked Jul 20 at 21:58
Answer Lee
48838
48838
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Perhaps you should more explicitly show that $f|_Y$ is bounded:
$$|f|_Y|_Y^* = sup_ = 1 |f|_Y(y)| = sup_ = 1 |f(y)| le sup_y |f(y)| = |f|_X^*$$
answered Jul 20 at 22:04
mechanodroid
22.2k52041
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Perhaps you should more explicitly show that $f|_Y$ is bounded:
$$|f|_Y|_Y^* = sup_ = 1 |f|_Y(y)| = sup_ = 1 |f(y)| le sup_y |f(y)| = |f|_X^*$$
answered Jul 20 at 22:04
mechanodroid
22.2k52041
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
add a comment |Â
up vote
2
down vote
Perhaps you should more explicitly show that $f|_Y$ is bounded:
$$|f|_Y|_Y^* = sup_ = 1 |f|_Y(y)| = sup_ = 1 |f(y)| le sup_y |f(y)| = |f|_X^*$$
answered Jul 20 at 22:04
mechanodroid
22.2k52041
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Perhaps you should more explicitly show that $f|_Y$ is bounded:
$$|f|_Y|_Y^* = sup_ = 1 |f|_Y(y)| = sup_ = 1 |f(y)| le sup_y |f(y)| = |f|_X^*$$
answered Jul 20 at 22:04
mechanodroid
22.2k52041
Perhaps you should more explicitly show that $f|_Y$ is bounded:
$$|f|_Y|_Y^* = sup_ = 1 |f|_Y(y)| = sup_ = 1 |f(y)| le sup_y |f(y)| = |f|_X^*$$
answered Jul 20 at 22:04
mechanodroid
22.2k52041
answered Jul 20 at 22:04
mechanodroid
22.2k52041
answered Jul 20 at 22:04
mechanodroid
22.2k52041
answered Jul 20 at 22:04
mechanodroid
22.2k52041
22.2k52041
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
add a comment |Â
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
Thanks for your answer but he told me to fix the logic here. But I couldn't find any logic mistakes.
– Answer Lee
Jul 20 at 22:06
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
@AnswerLee Well, $f_1 = f_2 implies f_1|_Y = f_2|_Y$ is a bit unnecessary as it is clear.
– mechanodroid
Jul 20 at 22:09
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
But I don't think it is a logic error. May be just redundant.
– Answer Lee
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
@AnswerLee Yes, that's what I meant.
– mechanodroid
Jul 20 at 22:11
add a comment |Â
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