Mixing
Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.
$begingroup$
Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.
I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this
I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.
Where i have;
$(x,y) = (y,x)$ for all $x,y in V$
and
$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector
I assumed it would simply be plugging in but im not entirely sure.
linear-algebra vector-spaces inner-product-space
edited Jan 7 at 14:14
Christoph
11.9k1642
asked Jan 7 at 13:18
L GL G
248
$endgroup$
add a comment |
$begingroup$
Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.
I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this
I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.
Where i have;
$(x,y) = (y,x)$ for all $x,y in V$
and
$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector
I assumed it would simply be plugging in but im not entirely sure.
linear-algebra vector-spaces inner-product-space
edited Jan 7 at 14:14
Christoph
11.9k1642
asked Jan 7 at 13:18
L GL G
248
$endgroup$
add a comment |
$begingroup$
Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.
I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this
I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.
Where i have;
$(x,y) = (y,x)$ for all $x,y in V$
and
$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector
I assumed it would simply be plugging in but im not entirely sure.
linear-algebra vector-spaces inner-product-space
edited Jan 7 at 14:14
Christoph
11.9k1642
asked Jan 7 at 13:18
L GL G
248
$endgroup$
Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.
I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this
I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.
Where i have;
$(x,y) = (y,x)$ for all $x,y in V$
and
$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector
I assumed it would simply be plugging in but im not entirely sure.
linear-algebra vector-spaces inner-product-space
linear-algebra vector-spaces inner-product-space
edited Jan 7 at 14:14
Christoph
11.9k1642
asked Jan 7 at 13:18
L GL G
248
edited Jan 7 at 14:14
Christoph
11.9k1642
asked Jan 7 at 13:18
L GL G
248
edited Jan 7 at 14:14
Christoph
11.9k1642
edited Jan 7 at 14:14
Christoph
11.9k1642
edited Jan 7 at 14:14
Christoph
11.9k1642
11.9k1642
asked Jan 7 at 13:18
L GL G
248
asked Jan 7 at 13:18
L GL G
248
asked Jan 7 at 13:18
L GL G
248
248
add a comment |
add a comment |
1 Answer
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$begingroup$
We have
$$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
And we have
$$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
for every f by $(f(t))^2geq 0$ and the monotony of the integral.
We now have to show that $(f,f)=0$ implies $f=0$.
It is standard lemma that for a continuous function $ggeq 0$ we have
$$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
$$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
If you don't know the lemma, I can give you a prove for it.
answered Jan 7 at 13:21
Student7Student7
2089
$endgroup$
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have
$$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
And we have
$$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
for every f by $(f(t))^2geq 0$ and the monotony of the integral.
We now have to show that $(f,f)=0$ implies $f=0$.
It is standard lemma that for a continuous function $ggeq 0$ we have
$$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
$$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
If you don't know the lemma, I can give you a prove for it.
answered Jan 7 at 13:21
Student7Student7
2089
$endgroup$
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
add a comment |
$begingroup$
We have
$$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
And we have
$$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
for every f by $(f(t))^2geq 0$ and the monotony of the integral.
We now have to show that $(f,f)=0$ implies $f=0$.
It is standard lemma that for a continuous function $ggeq 0$ we have
$$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
$$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
If you don't know the lemma, I can give you a prove for it.
answered Jan 7 at 13:21
Student7Student7
2089
$endgroup$
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
add a comment |
$begingroup$
We have
$$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
And we have
$$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
for every f by $(f(t))^2geq 0$ and the monotony of the integral.
We now have to show that $(f,f)=0$ implies $f=0$.
It is standard lemma that for a continuous function $ggeq 0$ we have
$$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
$$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
If you don't know the lemma, I can give you a prove for it.
answered Jan 7 at 13:21
Student7Student7
2089
$endgroup$
We have
$$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
And we have
$$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
for every f by $(f(t))^2geq 0$ and the monotony of the integral.
We now have to show that $(f,f)=0$ implies $f=0$.
It is standard lemma that for a continuous function $ggeq 0$ we have
$$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
$$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
If you don't know the lemma, I can give you a prove for it.
answered Jan 7 at 13:21
Student7Student7
2089
edited Jan 7 at 13:48
answered Jan 7 at 13:21
Student7Student7
2089
answered Jan 7 at 13:21
Student7Student7
2089
answered Jan 7 at 13:21
Student7Student7
2089
2089
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
add a comment |
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
$begingroup$
ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
$endgroup$
– L G
Jan 7 at 13:26
1
1
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
$begingroup$
+1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
$endgroup$
– Theo Bendit
Jan 7 at 14:02
add a comment |
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