Showing that $exists f in X^*$ : $|f| = frac{1}{d(x_0,Y)}, ; f(x_0) = 1$ and $f(y) = 0$.
$begingroup$
Exercise :
Let $X$ be a normed space and $Y$ be a proper closed subspace of $X$. If $x_0 notin Y$, show that there exists $f in X^*$ such that :
$$|f| = frac{1}{d(x_0,Y)}, ; f(x_0) = 1 ; text{and} ; f(y) = 0$$
Attempt :
I know the following Lemma :
Lemma : Let $(X,|cdot|)$ be a normed space and $Y$ be a proper closed subspace of $X$ with $x_0 in X setminus Y$. Then, there exists $f in S_{X^*}$ such that $f(y) =0$ and $f(x_0) = d(x_0,Y)$.
Using the lemma above for our exercise, I can say that there exists $g in S_{X^*}$ such that $g(x_0) = d(x_0,Y)$ and $g(y)=0$.
Now, let :
$$f := frac{1}{d(x_0,Y)}g$$
This means that :
$$f(x_0) = frac{1}{d(x_0,Y)}g(x_0) = 1 quad text{and} quad f(y) = 0$$
Finally, for the operator norm of $f$ it is $|g| = 1$ since $g in S_{X^*}$, which straightforwardly implies :
$$|f| = frac{1}{d(x_0,Y)}$$
Question : Is my approach correct and rigorous enough ? Other than that easy way out, is there a more standard solution approach ?
functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Exercise :
Let $X$ be a normed space and $Y$ be a proper closed subspace of $X$. If $x_0 notin Y$, show that there exists $f in X^*$ such that :
$$|f| = frac{1}{d(x_0,Y)}, ; f(x_0) = 1 ; text{and} ; f(y) = 0$$
Attempt :
I know the following Lemma :
Lemma : Let $(X,|cdot|)$ be a normed space and $Y$ be a proper closed subspace of $X$ with $x_0 in X setminus Y$. Then, there exists $f in S_{X^*}$ such that $f(y) =0$ and $f(x_0) = d(x_0,Y)$.
Using the lemma above for our exercise, I can say that there exists $g in S_{X^*}$ such that $g(x_0) = d(x_0,Y)$ and $g(y)=0$.
Now, let :
$$f := frac{1}{d(x_0,Y)}g$$
This means that :
$$f(x_0) = frac{1}{d(x_0,Y)}g(x_0) = 1 quad text{and} quad f(y) = 0$$
Finally, for the operator norm of $f$ it is $|g| = 1$ since $g in S_{X^*}$, which straightforwardly implies :
$$|f| = frac{1}{d(x_0,Y)}$$
Question : Is my approach correct and rigorous enough ? Other than that easy way out, is there a more standard solution approach ?
functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Exercise :
Let $X$ be a normed space and $Y$ be a proper closed subspace of $X$. If $x_0 notin Y$, show that there exists $f in X^*$ such that :
$$|f| = frac{1}{d(x_0,Y)}, ; f(x_0) = 1 ; text{and} ; f(y) = 0$$
Attempt :
I know the following Lemma :
Lemma : Let $(X,|cdot|)$ be a normed space and $Y$ be a proper closed subspace of $X$ with $x_0 in X setminus Y$. Then, there exists $f in S_{X^*}$ such that $f(y) =0$ and $f(x_0) = d(x_0,Y)$.
Using the lemma above for our exercise, I can say that there exists $g in S_{X^*}$ such that $g(x_0) = d(x_0,Y)$ and $g(y)=0$.
Now, let :
$$f := frac{1}{d(x_0,Y)}g$$
This means that :
$$f(x_0) = frac{1}{d(x_0,Y)}g(x_0) = 1 quad text{and} quad f(y) = 0$$
Finally, for the operator norm of $f$ it is $|g| = 1$ since $g in S_{X^*}$, which straightforwardly implies :
$$|f| = frac{1}{d(x_0,Y)}$$
Question : Is my approach correct and rigorous enough ? Other than that easy way out, is there a more standard solution approach ?
functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
$endgroup$
Exercise :
Let $X$ be a normed space and $Y$ be a proper closed subspace of $X$. If $x_0 notin Y$, show that there exists $f in X^*$ such that :
$$|f| = frac{1}{d(x_0,Y)}, ; f(x_0) = 1 ; text{and} ; f(y) = 0$$
Attempt :
I know the following Lemma :
Lemma : Let $(X,|cdot|)$ be a normed space and $Y$ be a proper closed subspace of $X$ with $x_0 in X setminus Y$. Then, there exists $f in S_{X^*}$ such that $f(y) =0$ and $f(x_0) = d(x_0,Y)$.
Using the lemma above for our exercise, I can say that there exists $g in S_{X^*}$ such that $g(x_0) = d(x_0,Y)$ and $g(y)=0$.
Now, let :
$$f := frac{1}{d(x_0,Y)}g$$
This means that :
$$f(x_0) = frac{1}{d(x_0,Y)}g(x_0) = 1 quad text{and} quad f(y) = 0$$
Finally, for the operator norm of $f$ it is $|g| = 1$ since $g in S_{X^*}$, which straightforwardly implies :
$$|f| = frac{1}{d(x_0,Y)}$$
Question : Is my approach correct and rigorous enough ? Other than that easy way out, is there a more standard solution approach ?
functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem
functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
asked Dec 17 '18 at 15:20
RebellosRebellos
14.8k31248
14.8k31248
add a comment |
add a comment |
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