If $Lambda$ is continuous at one point in $X$, then $Lambda$ is bounded on $X$.
This is taken from "Real and Complex Analysis" by Walter Rudin:
Theorem 5.4 For a linear transformation $Lambda$ of a normed linear space $X$ into a normed linear space $Y$, each of the following three conditions implies the other two:
(a) $Lambda$ is bounded
(b) $Lambda$ is continuous
(c) $Lambda$ is continuous at one point of $X$.
Proof: Since $|Lambda(x_1-x_2)| le |Lambda| |x_1-x_2|$, it is clear that (a) implies (b), and (b) implies (c) trivially. Suppose $Lambda$ is continuous at $x_0$. To each $epsilon > 0$, one can then find $delta > 0$ so that $|x-x_0|<delta$ implies $|Lambda x - Lambda x_0|<epsilon$. In other words, $|x|<delta$ implies $$|Lambda(x_0+x)-Lambda x_0|<epsilon.$$ But then the linearity of $Lambda$ shows that $|Lambda x|<epsilon$. Hence $|Lambda| le frac{epsilon}{delta}$, and (c) implies (a).
My question concerns only (c) implies (a). The supremum norm is $$|Lambda|=sup_{x not=0} frac{|Lambda x|}{|x|}.$$
However, I am not sure how $|Lambda x|<epsilon$ and $|x|<delta$ (which implies $frac 1{|x|} > frac 1{delta}$) imply $frac{|Lambda x|}{|x|} le epsilon$, since the opposing inequalities are confusing me.
real-analysis
asked Jul 22 '15 at 17:48
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This is taken from "Real and Complex Analysis" by Walter Rudin:
Theorem 5.4 For a linear transformation $Lambda$ of a normed linear space $X$ into a normed linear space $Y$, each of the following three conditions implies the other two:
(a) $Lambda$ is bounded
(b) $Lambda$ is continuous
(c) $Lambda$ is continuous at one point of $X$.
Proof: Since $|Lambda(x_1-x_2)| le |Lambda| |x_1-x_2|$, it is clear that (a) implies (b), and (b) implies (c) trivially. Suppose $Lambda$ is continuous at $x_0$. To each $epsilon > 0$, one can then find $delta > 0$ so that $|x-x_0|<delta$ implies $|Lambda x - Lambda x_0|<epsilon$. In other words, $|x|<delta$ implies $$|Lambda(x_0+x)-Lambda x_0|<epsilon.$$ But then the linearity of $Lambda$ shows that $|Lambda x|<epsilon$. Hence $|Lambda| le frac{epsilon}{delta}$, and (c) implies (a).
My question concerns only (c) implies (a). The supremum norm is $$|Lambda|=sup_{x not=0} frac{|Lambda x|}{|x|}.$$
However, I am not sure how $|Lambda x|<epsilon$ and $|x|<delta$ (which implies $frac 1{|x|} > frac 1{delta}$) imply $frac{|Lambda x|}{|x|} le epsilon$, since the opposing inequalities are confusing me.
real-analysis
asked Jul 22 '15 at 17:48
Cookie
8,684123580
add a comment |
This is taken from "Real and Complex Analysis" by Walter Rudin:
Theorem 5.4 For a linear transformation $Lambda$ of a normed linear space $X$ into a normed linear space $Y$, each of the following three conditions implies the other two:
(a) $Lambda$ is bounded
(b) $Lambda$ is continuous
(c) $Lambda$ is continuous at one point of $X$.
Proof: Since $|Lambda(x_1-x_2)| le |Lambda| |x_1-x_2|$, it is clear that (a) implies (b), and (b) implies (c) trivially. Suppose $Lambda$ is continuous at $x_0$. To each $epsilon > 0$, one can then find $delta > 0$ so that $|x-x_0|<delta$ implies $|Lambda x - Lambda x_0|<epsilon$. In other words, $|x|<delta$ implies $$|Lambda(x_0+x)-Lambda x_0|<epsilon.$$ But then the linearity of $Lambda$ shows that $|Lambda x|<epsilon$. Hence $|Lambda| le frac{epsilon}{delta}$, and (c) implies (a).
My question concerns only (c) implies (a). The supremum norm is $$|Lambda|=sup_{x not=0} frac{|Lambda x|}{|x|}.$$
However, I am not sure how $|Lambda x|<epsilon$ and $|x|<delta$ (which implies $frac 1{|x|} > frac 1{delta}$) imply $frac{|Lambda x|}{|x|} le epsilon$, since the opposing inequalities are confusing me.
real-analysis
asked Jul 22 '15 at 17:48
Cookie
8,684123580
This is taken from "Real and Complex Analysis" by Walter Rudin:
Theorem 5.4 For a linear transformation $Lambda$ of a normed linear space $X$ into a normed linear space $Y$, each of the following three conditions implies the other two:
(a) $Lambda$ is bounded
(b) $Lambda$ is continuous
(c) $Lambda$ is continuous at one point of $X$.
Proof: Since $|Lambda(x_1-x_2)| le |Lambda| |x_1-x_2|$, it is clear that (a) implies (b), and (b) implies (c) trivially. Suppose $Lambda$ is continuous at $x_0$. To each $epsilon > 0$, one can then find $delta > 0$ so that $|x-x_0|<delta$ implies $|Lambda x - Lambda x_0|<epsilon$. In other words, $|x|<delta$ implies $$|Lambda(x_0+x)-Lambda x_0|<epsilon.$$ But then the linearity of $Lambda$ shows that $|Lambda x|<epsilon$. Hence $|Lambda| le frac{epsilon}{delta}$, and (c) implies (a).
My question concerns only (c) implies (a). The supremum norm is $$|Lambda|=sup_{x not=0} frac{|Lambda x|}{|x|}.$$
However, I am not sure how $|Lambda x|<epsilon$ and $|x|<delta$ (which implies $frac 1{|x|} > frac 1{delta}$) imply $frac{|Lambda x|}{|x|} le epsilon$, since the opposing inequalities are confusing me.
real-analysis
real-analysis
asked Jul 22 '15 at 17:48
Cookie
8,684123580
asked Jul 22 '15 at 17:48
Cookie
8,684123580
edited Jul 22 '15 at 17:53
asked Jul 22 '15 at 17:48
Cookie
8,684123580
asked Jul 22 '15 at 17:48
Cookie
8,684123580
asked Jul 22 '15 at 17:48
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8,684123580
8,684123580
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2 Answers
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Pick $0 < r < 1$. For $x$ with $lVert xrVert = rdelta$, you know $lVert Lambda xrVert < epsilon$. Hence
$$frac{lVertLambda xrVert}{lVert xrVert} < frac{varepsilon}{rdelta}$$
for these $x$. For an arbitrary $y neq 0$, write
$$y = frac{lVert yrVert}{rdelta}cdot underbrace{frac{rdelta}{lVert yrVert}y}_x,$$
where $lVert xrVert = rdelta$. Then by the homogeneity of the norm and $Lambda$, you have
$$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert} < frac{epsilon}{rdelta}.$$
Let $r to 1$ to obtain the inequality
$$lVertLambdarVert leqslant frac{epsilon}{delta}.$$
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
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$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert}$, How?
answered Dec 2 '18 at 3:16
Heisenberg
65
add a comment |
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2 Answers
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2 Answers
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Pick $0 < r < 1$. For $x$ with $lVert xrVert = rdelta$, you know $lVert Lambda xrVert < epsilon$. Hence
$$frac{lVertLambda xrVert}{lVert xrVert} < frac{varepsilon}{rdelta}$$
for these $x$. For an arbitrary $y neq 0$, write
$$y = frac{lVert yrVert}{rdelta}cdot underbrace{frac{rdelta}{lVert yrVert}y}_x,$$
where $lVert xrVert = rdelta$. Then by the homogeneity of the norm and $Lambda$, you have
$$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert} < frac{epsilon}{rdelta}.$$
Let $r to 1$ to obtain the inequality
$$lVertLambdarVert leqslant frac{epsilon}{delta}.$$
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
add a comment |
Pick $0 < r < 1$. For $x$ with $lVert xrVert = rdelta$, you know $lVert Lambda xrVert < epsilon$. Hence
$$frac{lVertLambda xrVert}{lVert xrVert} < frac{varepsilon}{rdelta}$$
for these $x$. For an arbitrary $y neq 0$, write
$$y = frac{lVert yrVert}{rdelta}cdot underbrace{frac{rdelta}{lVert yrVert}y}_x,$$
where $lVert xrVert = rdelta$. Then by the homogeneity of the norm and $Lambda$, you have
$$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert} < frac{epsilon}{rdelta}.$$
Let $r to 1$ to obtain the inequality
$$lVertLambdarVert leqslant frac{epsilon}{delta}.$$
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
add a comment |
Pick $0 < r < 1$. For $x$ with $lVert xrVert = rdelta$, you know $lVert Lambda xrVert < epsilon$. Hence
$$frac{lVertLambda xrVert}{lVert xrVert} < frac{varepsilon}{rdelta}$$
for these $x$. For an arbitrary $y neq 0$, write
$$y = frac{lVert yrVert}{rdelta}cdot underbrace{frac{rdelta}{lVert yrVert}y}_x,$$
where $lVert xrVert = rdelta$. Then by the homogeneity of the norm and $Lambda$, you have
$$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert} < frac{epsilon}{rdelta}.$$
Let $r to 1$ to obtain the inequality
$$lVertLambdarVert leqslant frac{epsilon}{delta}.$$
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
Pick $0 < r < 1$. For $x$ with $lVert xrVert = rdelta$, you know $lVert Lambda xrVert < epsilon$. Hence
$$frac{lVertLambda xrVert}{lVert xrVert} < frac{varepsilon}{rdelta}$$
for these $x$. For an arbitrary $y neq 0$, write
$$y = frac{lVert yrVert}{rdelta}cdot underbrace{frac{rdelta}{lVert yrVert}y}_x,$$
where $lVert xrVert = rdelta$. Then by the homogeneity of the norm and $Lambda$, you have
$$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert} < frac{epsilon}{rdelta}.$$
Let $r to 1$ to obtain the inequality
$$lVertLambdarVert leqslant frac{epsilon}{delta}.$$
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
answered Jul 22 '15 at 17:57
Daniel Fischer♦
173k16160282
173k16160282
add a comment |
add a comment |
$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert}$, How?
answered Dec 2 '18 at 3:16
Heisenberg
65
add a comment |
$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert}$, How?
answered Dec 2 '18 at 3:16
Heisenberg
65
add a comment |
$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert}$, How?
answered Dec 2 '18 at 3:16
Heisenberg
65
$frac{lVert Lambda yrVert}{lVert yrVert} = frac{lVert Lambda xrVert}{lVert xrVert}$, How?
answered Dec 2 '18 at 3:16
Heisenberg
65
edited Dec 2 '18 at 3:22
answered Dec 2 '18 at 3:16
Heisenberg
65
answered Dec 2 '18 at 3:16
Heisenberg
65
answered Dec 2 '18 at 3:16
Heisenberg
65
65
add a comment |
add a comment |
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