Mixing
How to show positive definiteness?
$begingroup$
For continuous functions $f : [0, 1] rightarrow mathbb{R}$ and $g : [0, 1] rightarrow mathbb{R}$, define
$$langle f, grangle = int_{0}^{1} f(x)g(x)mathop{dx}.$$
How can I show this form is positive definite? I want to show
$$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$
with equality only if $f(x) = 0$. I know how to do this by assuming $f$ is differentiable, but is it possible without knowing this fact?
linear-algebra
$endgroup$
add a comment |
$begingroup$
For continuous functions $f : [0, 1] rightarrow mathbb{R}$ and $g : [0, 1] rightarrow mathbb{R}$, define
$$langle f, grangle = int_{0}^{1} f(x)g(x)mathop{dx}.$$
How can I show this form is positive definite? I want to show
$$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$
with equality only if $f(x) = 0$. I know how to do this by assuming $f$ is differentiable, but is it possible without knowing this fact?
linear-algebra
$endgroup$
add a comment |
$begingroup$
For continuous functions $f : [0, 1] rightarrow mathbb{R}$ and $g : [0, 1] rightarrow mathbb{R}$, define
$$langle f, grangle = int_{0}^{1} f(x)g(x)mathop{dx}.$$
How can I show this form is positive definite? I want to show
$$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$
with equality only if $f(x) = 0$. I know how to do this by assuming $f$ is differentiable, but is it possible without knowing this fact?
linear-algebra
$endgroup$
For continuous functions $f : [0, 1] rightarrow mathbb{R}$ and $g : [0, 1] rightarrow mathbb{R}$, define
$$langle f, grangle = int_{0}^{1} f(x)g(x)mathop{dx}.$$
How can I show this form is positive definite? I want to show
$$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$
with equality only if $f(x) = 0$. I know how to do this by assuming $f$ is differentiable, but is it possible without knowing this fact?
linear-algebra
linear-algebra
asked Jan 29 at 23:36
user614735
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, $$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$ should be clear from the definition of the integral.
Now assume $f$ is not identically zero. Then there is an $x_0in [0,1]$ with $delta:=f(x_0)>0.$ As $f$ is continuous, there is an $varepsilon>0$ such that $f(x)>delta/2$ for all $xin [0,1]$ with $|x-x_0|<varepsilon.$ Now the integral has to be at least $varepsilon cdot delta^2/4>0.$
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
add a comment |
$begingroup$
So, the theorem you want:
Let $g$ be a nonnegative continuous function on $[a,b]$. Suppose that there is some $xin [a,b]$ such that $g(x) > 0$. Then $int_a^b g(t),dt > 0$.
Proof: Put a box under the graph:
The integral is at least the area of the box, which is positive. Why does a box fit? Because if we choose $epsilon$ between $0$ and $g(x)$, there is some $delta$ such that $g(x+y)>g(x)-epsilon$ for $|y|<delta$. The box runs from $x-delta$ to $x+delta$ horizontally, and up to $g(x)-epsilon$ vertically.
Addendum: we didn't actually use global continuity, only continuity at the single point $x$. As such, several variations of this theorem are possible, such as the following: if $f$ is integrable and strictly positive on an interval $I$, and $f$ is continuous at at least one point of $I$, then $int_I f > 0$.
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
First, $$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$ should be clear from the definition of the integral.
Now assume $f$ is not identically zero. Then there is an $x_0in [0,1]$ with $delta:=f(x_0)>0.$ As $f$ is continuous, there is an $varepsilon>0$ such that $f(x)>delta/2$ for all $xin [0,1]$ with $|x-x_0|<varepsilon.$ Now the integral has to be at least $varepsilon cdot delta^2/4>0.$
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
add a comment |
$begingroup$
First, $$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$ should be clear from the definition of the integral.
Now assume $f$ is not identically zero. Then there is an $x_0in [0,1]$ with $delta:=f(x_0)>0.$ As $f$ is continuous, there is an $varepsilon>0$ such that $f(x)>delta/2$ for all $xin [0,1]$ with $|x-x_0|<varepsilon.$ Now the integral has to be at least $varepsilon cdot delta^2/4>0.$
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
add a comment |
$begingroup$
First, $$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$ should be clear from the definition of the integral.
Now assume $f$ is not identically zero. Then there is an $x_0in [0,1]$ with $delta:=f(x_0)>0.$ As $f$ is continuous, there is an $varepsilon>0$ such that $f(x)>delta/2$ for all $xin [0,1]$ with $|x-x_0|<varepsilon.$ Now the integral has to be at least $varepsilon cdot delta^2/4>0.$
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
$endgroup$
First, $$int_{0}^{1} f(x)^{2} mathop{dx} geq 0 $$ should be clear from the definition of the integral.
Now assume $f$ is not identically zero. Then there is an $x_0in [0,1]$ with $delta:=f(x_0)>0.$ As $f$ is continuous, there is an $varepsilon>0$ such that $f(x)>delta/2$ for all $xin [0,1]$ with $|x-x_0|<varepsilon.$ Now the integral has to be at least $varepsilon cdot delta^2/4>0.$
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
edited Feb 2 at 9:49
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 23:40
Reiner MartinReiner Martin
3,509414
3,509414
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
add a comment |
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
why must the integral be at least $epsilon cdot delta/2$? I thought it must be at least $(delta/2)$^{2}$?
$endgroup$
– user614735
Feb 2 at 4:49
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
Correct, I'll make a correction. Doesn't change the core of the argument to be sure. Thanks for spotting this
$endgroup$
– Reiner Martin
Feb 2 at 9:48
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
okay. why is it $epsilon cdot delta^{2}/4$ and not just $delta^{2}/4$? because you can do $int_{0}^{1} f(x)^{2} mathop{dx} geq int_{0}^{1}delta^{2}/4 mathop{dx} = delta^{2}/4 int_{0}^{1} mathop{dx} = delta^{2}/4$
$endgroup$
– user614735
Feb 2 at 15:08
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
You can't do that as you only know that $f(x)>delta/2$ for those $x$ in the interval $(x_0-varepsilon,x_0+varepsilon) cap [0,1].$ That interval has length at least $varepsilon,$ however.
$endgroup$
– Reiner Martin
Feb 2 at 23:25
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
$begingroup$
ok thanks it makes sense
$endgroup$
– user614735
Feb 3 at 2:48
add a comment |
$begingroup$
So, the theorem you want:
Let $g$ be a nonnegative continuous function on $[a,b]$. Suppose that there is some $xin [a,b]$ such that $g(x) > 0$. Then $int_a^b g(t),dt > 0$.
Proof: Put a box under the graph:
The integral is at least the area of the box, which is positive. Why does a box fit? Because if we choose $epsilon$ between $0$ and $g(x)$, there is some $delta$ such that $g(x+y)>g(x)-epsilon$ for $|y|<delta$. The box runs from $x-delta$ to $x+delta$ horizontally, and up to $g(x)-epsilon$ vertically.
Addendum: we didn't actually use global continuity, only continuity at the single point $x$. As such, several variations of this theorem are possible, such as the following: if $f$ is integrable and strictly positive on an interval $I$, and $f$ is continuous at at least one point of $I$, then $int_I f > 0$.
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
So, the theorem you want:
Let $g$ be a nonnegative continuous function on $[a,b]$. Suppose that there is some $xin [a,b]$ such that $g(x) > 0$. Then $int_a^b g(t),dt > 0$.
Proof: Put a box under the graph:
The integral is at least the area of the box, which is positive. Why does a box fit? Because if we choose $epsilon$ between $0$ and $g(x)$, there is some $delta$ such that $g(x+y)>g(x)-epsilon$ for $|y|<delta$. The box runs from $x-delta$ to $x+delta$ horizontally, and up to $g(x)-epsilon$ vertically.
Addendum: we didn't actually use global continuity, only continuity at the single point $x$. As such, several variations of this theorem are possible, such as the following: if $f$ is integrable and strictly positive on an interval $I$, and $f$ is continuous at at least one point of $I$, then $int_I f > 0$.
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
So, the theorem you want:
Let $g$ be a nonnegative continuous function on $[a,b]$. Suppose that there is some $xin [a,b]$ such that $g(x) > 0$. Then $int_a^b g(t),dt > 0$.
Proof: Put a box under the graph:
The integral is at least the area of the box, which is positive. Why does a box fit? Because if we choose $epsilon$ between $0$ and $g(x)$, there is some $delta$ such that $g(x+y)>g(x)-epsilon$ for $|y|<delta$. The box runs from $x-delta$ to $x+delta$ horizontally, and up to $g(x)-epsilon$ vertically.
Addendum: we didn't actually use global continuity, only continuity at the single point $x$. As such, several variations of this theorem are possible, such as the following: if $f$ is integrable and strictly positive on an interval $I$, and $f$ is continuous at at least one point of $I$, then $int_I f > 0$.
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
$endgroup$
So, the theorem you want:
Let $g$ be a nonnegative continuous function on $[a,b]$. Suppose that there is some $xin [a,b]$ such that $g(x) > 0$. Then $int_a^b g(t),dt > 0$.
Proof: Put a box under the graph:
The integral is at least the area of the box, which is positive. Why does a box fit? Because if we choose $epsilon$ between $0$ and $g(x)$, there is some $delta$ such that $g(x+y)>g(x)-epsilon$ for $|y|<delta$. The box runs from $x-delta$ to $x+delta$ horizontally, and up to $g(x)-epsilon$ vertically.
Addendum: we didn't actually use global continuity, only continuity at the single point $x$. As such, several variations of this theorem are possible, such as the following: if $f$ is integrable and strictly positive on an interval $I$, and $f$ is continuous at at least one point of $I$, then $int_I f > 0$.
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
edited Jan 30 at 6:28
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
answered Jan 29 at 23:54
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
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