2022-07-09

We define πΆ[π, π] as the set of all continuous functions π:[π, π] β β. Let π = απ β πΆ[π, π]: β« π(π₯)ππ₯ ΰ― ΰ― = 0α. Show that π is a subspace of πΆ[π, π].

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To show that π is a subspace of πΆ[π, π], we need to prove three properties:
1. π is closed under addition.
2. π is closed under scalar multiplication.
3. π contains the zero function.
Let's go through each of these properties:
1. π is closed under addition:
To show that π is closed under addition, we need to prove that if πβ, πβ β π, then πβ + πβ also belongs to π.
Let πβ, πβ β π, which means β«(πβ(π₯))ππ₯ = 0 and β«(πβ(π₯))ππ₯ = 0.
We want to show that β«((πβ + πβ)(π₯))ππ₯ = 0.
Using the linearity of the integral, we have:
β«((πβ + πβ)(π₯))ππ₯ = β«(πβ(π₯) + πβ(π₯))ππ₯.
Now, let's compute this integral:
$\beta «\left({f}_{1}\left(x\right)+{f}_{2}\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx=\beta «{f}_{1}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx+\beta «{f}_{2}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx.$
Since β«(πβ(π₯))ππ₯ = 0 and β«(πβ(π₯))ππ₯ = 0, we can substitute these values:
$\beta «\left({f}_{1}\left(x\right)+{f}_{2}\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx=0+0=0.$
Therefore, πβ + πβ also belongs to π, and π is closed under addition.
2. π is closed under scalar multiplication:
To show that π is closed under scalar multiplication, we need to prove that if π β π and π is a scalar, then ππ also belongs to π.
Let π β π, which means β«(π(π₯))ππ₯ = 0.
We want to show that β«((ππ)(π₯))ππ₯ = 0.
Using the linearity of the integral, we have:
β«((ππ)(π₯))ππ₯ = πβ«(π(π₯))ππ₯.
Now, let's compute this integral:
$\beta «\left(c\mathrm{Β·}f\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx=c\mathrm{Β·}\beta «f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx.$
Since β«(π(π₯))ππ₯ = 0, we can substitute this value:
$\beta «\left(c\mathrm{Β·}f\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx=c\mathrm{Β·}0=0.$
Therefore, ππ also belongs to π, and π is closed under scalar multiplication.
3. π contains the zero function:
To show that π contains the zero function, we need to prove that the zero function π(π₯) = 0 belongs to π.
To do this, we need to compute the integral:
$\beta «f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\beta «0\phantom{\rule{0.167em}{0ex}}dx=0.$
Since the integral of the zero function over any interval is zero, the zero function π(π₯) = 0 belongs to π.
Therefore, π satisfies all three properties: it is closed under addition, closed under scalar multiplication, and contains the zero function. Hence, π is a subspace of πΆ[π, π].

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