Bounded Linear Functional and Riesz's Lemma
$begingroup$
Let $ X = C[0,1] $ with the supremum norm $ || cdot ||_infty $. Consider the functional $ phi $ defined for $ f in X $ by
$$ phi(f) = int_0^frac{1}{2} f(x) dx - int_frac{1}{2}^1 f(x) dx .$$
Show that if $ f in X $ with $ ||f|| = 1 $ then $ |phi(f)| < 1 $.
A previous part of this question was to show that $ phi $ is a bounded linear functional with norm 1 which I have done already. But I do not know how to show that $ |phi(f)| < 1 $. Any help or hints would be appreciated.
functional-analysis normed-spaces
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
$endgroup$
add a comment |
$begingroup$
Let $ X = C[0,1] $ with the supremum norm $ || cdot ||_infty $. Consider the functional $ phi $ defined for $ f in X $ by
$$ phi(f) = int_0^frac{1}{2} f(x) dx - int_frac{1}{2}^1 f(x) dx .$$
Show that if $ f in X $ with $ ||f|| = 1 $ then $ |phi(f)| < 1 $.
A previous part of this question was to show that $ phi $ is a bounded linear functional with norm 1 which I have done already. But I do not know how to show that $ |phi(f)| < 1 $. Any help or hints would be appreciated.
functional-analysis normed-spaces
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
$endgroup$
add a comment |
$begingroup$
Let $ X = C[0,1] $ with the supremum norm $ || cdot ||_infty $. Consider the functional $ phi $ defined for $ f in X $ by
$$ phi(f) = int_0^frac{1}{2} f(x) dx - int_frac{1}{2}^1 f(x) dx .$$
Show that if $ f in X $ with $ ||f|| = 1 $ then $ |phi(f)| < 1 $.
A previous part of this question was to show that $ phi $ is a bounded linear functional with norm 1 which I have done already. But I do not know how to show that $ |phi(f)| < 1 $. Any help or hints would be appreciated.
functional-analysis normed-spaces
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
$endgroup$
Let $ X = C[0,1] $ with the supremum norm $ || cdot ||_infty $. Consider the functional $ phi $ defined for $ f in X $ by
$$ phi(f) = int_0^frac{1}{2} f(x) dx - int_frac{1}{2}^1 f(x) dx .$$
Show that if $ f in X $ with $ ||f|| = 1 $ then $ |phi(f)| < 1 $.
A previous part of this question was to show that $ phi $ is a bounded linear functional with norm 1 which I have done already. But I do not know how to show that $ |phi(f)| < 1 $. Any help or hints would be appreciated.
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
asked Nov 30 '18 at 23:12
Chase SariaslaniChase Sariaslani
805
805
add a comment |
add a comment |
1 Answer
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Prove by contradiction. Suppose $|phi (f)|=1$. Then $1=|int_0^{1/2}f-int_{1/2}^{1}f| leq frac 1 2 +frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f equiv 1$ or $f equiv -1$. But then $phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
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1 Answer
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1 Answer
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$begingroup$
Prove by contradiction. Suppose $|phi (f)|=1$. Then $1=|int_0^{1/2}f-int_{1/2}^{1}f| leq frac 1 2 +frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f equiv 1$ or $f equiv -1$. But then $phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
add a comment |
$begingroup$
Prove by contradiction. Suppose $|phi (f)|=1$. Then $1=|int_0^{1/2}f-int_{1/2}^{1}f| leq frac 1 2 +frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f equiv 1$ or $f equiv -1$. But then $phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
add a comment |
$begingroup$
Prove by contradiction. Suppose $|phi (f)|=1$. Then $1=|int_0^{1/2}f-int_{1/2}^{1}f| leq frac 1 2 +frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f equiv 1$ or $f equiv -1$. But then $phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
Prove by contradiction. Suppose $|phi (f)|=1$. Then $1=|int_0^{1/2}f-int_{1/2}^{1}f| leq frac 1 2 +frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f equiv 1$ or $f equiv -1$. But then $phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Nov 30 '18 at 23:20
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
52.9k32055
add a comment |
add a comment |
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