Vector subspaces of continuous functions
Let $V = C^0 [a, b]$ be the set of all real-valued continuous functions on the domain $[a, b]$. Which of the following subsets $W_1,W_2$ are subspaces of $V$?
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
Okay, I know to show that $W$ is a subspace of $V$:
a. $W$ is non-empty.
b. if $x_1, x_2 in W$ then $x_1 + x_2 in W$
c. for $k in R, kx_1 in W$
vector-spaces
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
bumped to the homepage by Community♦ 18 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
|
show 14 more comments
Let $V = C^0 [a, b]$ be the set of all real-valued continuous functions on the domain $[a, b]$. Which of the following subsets $W_1,W_2$ are subspaces of $V$?
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
Okay, I know to show that $W$ is a subspace of $V$:
a. $W$ is non-empty.
b. if $x_1, x_2 in W$ then $x_1 + x_2 in W$
c. for $k in R, kx_1 in W$
vector-spaces
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
bumped to the homepage by Community♦ 18 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
2
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41
|
show 14 more comments
Let $V = C^0 [a, b]$ be the set of all real-valued continuous functions on the domain $[a, b]$. Which of the following subsets $W_1,W_2$ are subspaces of $V$?
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
Okay, I know to show that $W$ is a subspace of $V$:
a. $W$ is non-empty.
b. if $x_1, x_2 in W$ then $x_1 + x_2 in W$
c. for $k in R, kx_1 in W$
vector-spaces
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
Let $V = C^0 [a, b]$ be the set of all real-valued continuous functions on the domain $[a, b]$. Which of the following subsets $W_1,W_2$ are subspaces of $V$?
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
Okay, I know to show that $W$ is a subspace of $V$:
a. $W$ is non-empty.
b. if $x_1, x_2 in W$ then $x_1 + x_2 in W$
c. for $k in R, kx_1 in W$
vector-spaces
vector-spaces
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
edited Oct 6 '13 at 23:04
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
asked Oct 6 '13 at 22:08
Zhoe
2,1261812
2,1261812
bumped to the homepage by Community♦ 18 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 18 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
2
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41
|
show 14 more comments
Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
2
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41
Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
2
2
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41
|
show 14 more comments
1 Answer
1
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oldest
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$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
a. $W_1$ is non- empty.
$f(x) = 0$
$|0| le M$
$0 le M$
since $M in R^+$ then $W_1$ is non-empty.
b. if $x_1, x_2 in W_1$ then $x_1 + x_2 in W_1$
$f(x), g(x) in W_1$
$f(x), g(x)$ are continuous and $|f(x)| le M, |g(x)| le N$
show that $h(x) in W_1$ is such that $h(x)$ is continuous and $|h(x)| le P$ for some $P in R^+$
if $f(x)$ and $g(x)$ are continuous at a, then $f(x) + g(x)$ are continuous at a.
$lim{xto_a} f(x) = f(a)$ and $lim{xto_a} g(x) = g(a)$
$lim{xto_a} f(x) + g(x) = lim{xto_a} f(x) + lim{xto_a} g(x) = f(a) + g(a)$, so $f(x) + g(x)$ is continuous.
let $h(x) = f(x) + g(x)$ and therefore $h(x)$ would be continuous.
if $|f(x)| le M, |g(x)| le N$ then $h(x) = |f(x) + g(x)| le M + N in R^+$
let $P = M + N$, therefore $|h(x)| = |f(x) + g(x)| le P in R^+$
since $h(x)$ is continuous and for some $P in R^+ |h(x)| le P, h(x) in W_1$ implies $f(x) + g(x) in W_1$
c. for $k in R, kx_1 in W_1$
similarly for $k in R$
$k cdot lim{xto_a} f(x) = k cdot f(a)$, is continuous at a.
thus, $k cdot f(x) le k cdot M in W_1$
by subspace test, $W_1$ is a subspace of $V$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
a. $W_2$ is non-empty
$f(x) = 0$
$f(-0) = f(0)$
the condition holds, thus $W_2$ is non-empty.
b. if $x_1, x_2 in W_2$ then $x_1 + x_2 in W_2$
$f(x), g(x) in W_2$
they are continuous and $f(-x) = f(x), g(-x) = g(x)$ for all x
this result means that $f$ and $g$ are even functions
$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$ for all x
hence, $f + g$ is also even. therefore $f + g in W_2$
c. for $k in R, kx_1 in W_2$
for $k in R$,
$(kf)(-x) = kf(-x) = k(f(-x)) = k(f(x)) = kf(x)$ for all x
hence $kf$ is even. therefore, $kf in W_2$
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
add a comment |
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$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
a. $W_1$ is non- empty.
$f(x) = 0$
$|0| le M$
$0 le M$
since $M in R^+$ then $W_1$ is non-empty.
b. if $x_1, x_2 in W_1$ then $x_1 + x_2 in W_1$
$f(x), g(x) in W_1$
$f(x), g(x)$ are continuous and $|f(x)| le M, |g(x)| le N$
show that $h(x) in W_1$ is such that $h(x)$ is continuous and $|h(x)| le P$ for some $P in R^+$
if $f(x)$ and $g(x)$ are continuous at a, then $f(x) + g(x)$ are continuous at a.
$lim{xto_a} f(x) = f(a)$ and $lim{xto_a} g(x) = g(a)$
$lim{xto_a} f(x) + g(x) = lim{xto_a} f(x) + lim{xto_a} g(x) = f(a) + g(a)$, so $f(x) + g(x)$ is continuous.
let $h(x) = f(x) + g(x)$ and therefore $h(x)$ would be continuous.
if $|f(x)| le M, |g(x)| le N$ then $h(x) = |f(x) + g(x)| le M + N in R^+$
let $P = M + N$, therefore $|h(x)| = |f(x) + g(x)| le P in R^+$
since $h(x)$ is continuous and for some $P in R^+ |h(x)| le P, h(x) in W_1$ implies $f(x) + g(x) in W_1$
c. for $k in R, kx_1 in W_1$
similarly for $k in R$
$k cdot lim{xto_a} f(x) = k cdot f(a)$, is continuous at a.
thus, $k cdot f(x) le k cdot M in W_1$
by subspace test, $W_1$ is a subspace of $V$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
a. $W_2$ is non-empty
$f(x) = 0$
$f(-0) = f(0)$
the condition holds, thus $W_2$ is non-empty.
b. if $x_1, x_2 in W_2$ then $x_1 + x_2 in W_2$
$f(x), g(x) in W_2$
they are continuous and $f(-x) = f(x), g(-x) = g(x)$ for all x
this result means that $f$ and $g$ are even functions
$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$ for all x
hence, $f + g$ is also even. therefore $f + g in W_2$
c. for $k in R, kx_1 in W_2$
for $k in R$,
$(kf)(-x) = kf(-x) = k(f(-x)) = k(f(x)) = kf(x)$ for all x
hence $kf$ is even. therefore, $kf in W_2$
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
add a comment |
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
a. $W_1$ is non- empty.
$f(x) = 0$
$|0| le M$
$0 le M$
since $M in R^+$ then $W_1$ is non-empty.
b. if $x_1, x_2 in W_1$ then $x_1 + x_2 in W_1$
$f(x), g(x) in W_1$
$f(x), g(x)$ are continuous and $|f(x)| le M, |g(x)| le N$
show that $h(x) in W_1$ is such that $h(x)$ is continuous and $|h(x)| le P$ for some $P in R^+$
if $f(x)$ and $g(x)$ are continuous at a, then $f(x) + g(x)$ are continuous at a.
$lim{xto_a} f(x) = f(a)$ and $lim{xto_a} g(x) = g(a)$
$lim{xto_a} f(x) + g(x) = lim{xto_a} f(x) + lim{xto_a} g(x) = f(a) + g(a)$, so $f(x) + g(x)$ is continuous.
let $h(x) = f(x) + g(x)$ and therefore $h(x)$ would be continuous.
if $|f(x)| le M, |g(x)| le N$ then $h(x) = |f(x) + g(x)| le M + N in R^+$
let $P = M + N$, therefore $|h(x)| = |f(x) + g(x)| le P in R^+$
since $h(x)$ is continuous and for some $P in R^+ |h(x)| le P, h(x) in W_1$ implies $f(x) + g(x) in W_1$
c. for $k in R, kx_1 in W_1$
similarly for $k in R$
$k cdot lim{xto_a} f(x) = k cdot f(a)$, is continuous at a.
thus, $k cdot f(x) le k cdot M in W_1$
by subspace test, $W_1$ is a subspace of $V$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
a. $W_2$ is non-empty
$f(x) = 0$
$f(-0) = f(0)$
the condition holds, thus $W_2$ is non-empty.
b. if $x_1, x_2 in W_2$ then $x_1 + x_2 in W_2$
$f(x), g(x) in W_2$
they are continuous and $f(-x) = f(x), g(-x) = g(x)$ for all x
this result means that $f$ and $g$ are even functions
$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$ for all x
hence, $f + g$ is also even. therefore $f + g in W_2$
c. for $k in R, kx_1 in W_2$
for $k in R$,
$(kf)(-x) = kf(-x) = k(f(-x)) = k(f(x)) = kf(x)$ for all x
hence $kf$ is even. therefore, $kf in W_2$
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
add a comment |
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
a. $W_1$ is non- empty.
$f(x) = 0$
$|0| le M$
$0 le M$
since $M in R^+$ then $W_1$ is non-empty.
b. if $x_1, x_2 in W_1$ then $x_1 + x_2 in W_1$
$f(x), g(x) in W_1$
$f(x), g(x)$ are continuous and $|f(x)| le M, |g(x)| le N$
show that $h(x) in W_1$ is such that $h(x)$ is continuous and $|h(x)| le P$ for some $P in R^+$
if $f(x)$ and $g(x)$ are continuous at a, then $f(x) + g(x)$ are continuous at a.
$lim{xto_a} f(x) = f(a)$ and $lim{xto_a} g(x) = g(a)$
$lim{xto_a} f(x) + g(x) = lim{xto_a} f(x) + lim{xto_a} g(x) = f(a) + g(a)$, so $f(x) + g(x)$ is continuous.
let $h(x) = f(x) + g(x)$ and therefore $h(x)$ would be continuous.
if $|f(x)| le M, |g(x)| le N$ then $h(x) = |f(x) + g(x)| le M + N in R^+$
let $P = M + N$, therefore $|h(x)| = |f(x) + g(x)| le P in R^+$
since $h(x)$ is continuous and for some $P in R^+ |h(x)| le P, h(x) in W_1$ implies $f(x) + g(x) in W_1$
c. for $k in R, kx_1 in W_1$
similarly for $k in R$
$k cdot lim{xto_a} f(x) = k cdot f(a)$, is continuous at a.
thus, $k cdot f(x) le k cdot M in W_1$
by subspace test, $W_1$ is a subspace of $V$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
a. $W_2$ is non-empty
$f(x) = 0$
$f(-0) = f(0)$
the condition holds, thus $W_2$ is non-empty.
b. if $x_1, x_2 in W_2$ then $x_1 + x_2 in W_2$
$f(x), g(x) in W_2$
they are continuous and $f(-x) = f(x), g(-x) = g(x)$ for all x
this result means that $f$ and $g$ are even functions
$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$ for all x
hence, $f + g$ is also even. therefore $f + g in W_2$
c. for $k in R, kx_1 in W_2$
for $k in R$,
$(kf)(-x) = kf(-x) = k(f(-x)) = k(f(x)) = kf(x)$ for all x
hence $kf$ is even. therefore, $kf in W_2$
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
$W_1 ={f in C^0 [a, b] : |f(x)| le M,, text{for some},, M in R^+}$, the set of bounded continuous functions on $[a, b]$
a. $W_1$ is non- empty.
$f(x) = 0$
$|0| le M$
$0 le M$
since $M in R^+$ then $W_1$ is non-empty.
b. if $x_1, x_2 in W_1$ then $x_1 + x_2 in W_1$
$f(x), g(x) in W_1$
$f(x), g(x)$ are continuous and $|f(x)| le M, |g(x)| le N$
show that $h(x) in W_1$ is such that $h(x)$ is continuous and $|h(x)| le P$ for some $P in R^+$
if $f(x)$ and $g(x)$ are continuous at a, then $f(x) + g(x)$ are continuous at a.
$lim{xto_a} f(x) = f(a)$ and $lim{xto_a} g(x) = g(a)$
$lim{xto_a} f(x) + g(x) = lim{xto_a} f(x) + lim{xto_a} g(x) = f(a) + g(a)$, so $f(x) + g(x)$ is continuous.
let $h(x) = f(x) + g(x)$ and therefore $h(x)$ would be continuous.
if $|f(x)| le M, |g(x)| le N$ then $h(x) = |f(x) + g(x)| le M + N in R^+$
let $P = M + N$, therefore $|h(x)| = |f(x) + g(x)| le P in R^+$
since $h(x)$ is continuous and for some $P in R^+ |h(x)| le P, h(x) in W_1$ implies $f(x) + g(x) in W_1$
c. for $k in R, kx_1 in W_1$
similarly for $k in R$
$k cdot lim{xto_a} f(x) = k cdot f(a)$, is continuous at a.
thus, $k cdot f(x) le k cdot M in W_1$
by subspace test, $W_1$ is a subspace of $V$
$W_2 = {f in C^0 [a, b] : f(-x) = f(x),, text{for all},, x}$, the set of even continuous functions on $[a, b]$
a. $W_2$ is non-empty
$f(x) = 0$
$f(-0) = f(0)$
the condition holds, thus $W_2$ is non-empty.
b. if $x_1, x_2 in W_2$ then $x_1 + x_2 in W_2$
$f(x), g(x) in W_2$
they are continuous and $f(-x) = f(x), g(-x) = g(x)$ for all x
this result means that $f$ and $g$ are even functions
$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$ for all x
hence, $f + g$ is also even. therefore $f + g in W_2$
c. for $k in R, kx_1 in W_2$
for $k in R$,
$(kf)(-x) = kf(-x) = k(f(-x)) = k(f(x)) = kf(x)$ for all x
hence $kf$ is even. therefore, $kf in W_2$
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
edited Oct 8 '13 at 6:41
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
answered Oct 8 '13 at 6:35
Zhoe
2,1261812
2,1261812
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Do you know the list of things that a subspace must satisfy?
– Joe Johnson 126
Oct 6 '13 at 22:11
yes I do, I'm just not sure how to go about showing it.
– Zhoe
Oct 6 '13 at 22:15
Well, since you know what you have to show, why not make a start on showing it, and we'll help you when you get stuck. Start by writing out what it is that you have to do to show something is a subspace.
– Gerry Myerson
Oct 6 '13 at 22:44
OK, good start. Now: can you show $W_1$ is non-empty?
– Gerry Myerson
Oct 6 '13 at 23:09
2
The question is ill posed. What are the unbounded continuous functions on $[a,b]$? And what's the meaning of “$f(-x)=f(x)$ for all $x$” if $[a,b]=[1,2]$?
– egreg
Oct 6 '13 at 23:41