Question

In \(C([0,1])\) find d(f,g) where f(x) =1 and g(x) =x for = \(\int_0^1f(x)g(x)dx\) Please, help

Answer 1

@SithsAndGiggles

Answer 2

Is \(d(f,g)\) supposed to be a metric?

Answer 3

Yes

Answer 4

It's distance function

Answer 5

Alright, so in this set, what is the metric defined as? Is it the same as the modulus?

Answer 6

d(v,w) =||v-w||

Answer 7

It's a norm

Answer 8

Okay, if I'm remembering this correctly, I think the norm of a vector is given by the square root of the inner product of the vector with itself.
\[\begin{align*}
d(f,g)&=\|f-g\|\\\\
&=\sqrt{\langle f-g,f-g\rangle}\\\\
&=\int_0^1\bigg(f(x)-g(x)\bigg)^2~dx\end{align*}\]

Answer 9

Oops, forgot the square root:
\[d(f,g)=\color{red}{\sqrt{\color{black}{\int_0^1\bigg(f(x)-g(x)\bigg)^2~dx}}}\]

Answer 10

\[\begin{align*}
d(f,g)&=\|f-g\|\\\\
&=\sqrt{\langle f-g,f-g\rangle}\\\\
&=\int_0^1\bigg(f(x)-g(x)\bigg)^2~dx\end{align*}\]
I don't get the last line

Answer 11

\[\sqrt {<f-g, f-g>}=\sqrt{<f, f-g>-<g,f-g>}\\=\sqrt{<f,f>-<g,f>-<g,f>+<g,g>}\]

Answer 12

\[=\sqrt{||f||^2-2\int_0^1f(x)g(x)dx +||g||^2}\]

Answer 13

So that I don't get how can you get the last line

Answer 14

I missed adding the square root. You should get the same thing when it's included in my equations.
\[\begin{align*}\sqrt{\int_0^1\bigg(f(x)-g(x)\bigg)^2~dx}&=\sqrt{\int_0^1\bigg(f^2(x)-2f(x)g(x)+g^2(x)\bigg)~dx}\\\\
&=\sqrt{\langle f,f\rangle-2\langle g,f\rangle+\langle g,g\rangle}
\end{align*}\]

Answer 15

oh, it's new to me since I don't have that equation :).

Answer 16

Yes, so it is 1, right?

Answer 17

I'm getting \(\dfrac{1}{\sqrt3}\)... Make sure you're integrating correctly.

Answer 18

show me please,

Answer 19

\[\begin{align*}
d(f,g)&=\|f-g\|\\\\
&=\sqrt{\langle f-g,f-g\rangle}\\\\
&=\sqrt{\int_0^1\bigg(f(x)-g(x)\bigg)^2~dx}\\\\
&=\sqrt{\int_0^1(1-x)^2~dx}\\\\
&=\sqrt{-\int_1^0u^2~du}\\\\
&=\sqrt{\int_0^1u^2~du}\\\\
&=\sqrt{\frac{1}{3}}
\end{align*}\]

Answer 20

I got you. Thanks a lot
How about sup norm?

Answer 21

By definition of sup norm , the metric given by sup norm is the largest vertical separation between the functions. Our functions have the interval [0,1], so that sup norm =1, right?

Answer 22

I think so, yes.

Answer 23

I still have problem on our different ways :(

Answer 24

I'm not seeing what's so different about them... From the defined inner product, you have
\[\langle f-g,f-g\rangle=\int_0^1(f-g)(f-g)~dx\]

Answer 25

\[\sqrt{<f,f>-<g,f>-<g,f>+<g,g>}=\sqrt{||f||^2-2<f,g>+||g||^2}\\=\sqrt{1-2\int_0^1xdx+x^2}\]

Answer 26

it's =1
but open on your way, it is \(1/sqrt 3\)

Answer 27

\[\sqrt{1-2\int_0^1x~dx+\int_0^1 x^2~dx}=\sqrt{1-2\left(\frac{1}{2}\right)+\frac{1}{3}}\]

Answer 28

Got you. I have to apply the definition on <g,g> also, not just ||g||^2 like what I did. right?

Answer 29

Yeah,
\[\langle g,g\rangle=\int_0^1x\cdot x~dx=\cdots=\|g\|^2\]

Answer 30

Thanks a lot. I do appreciate :)

Answer 31

You're welcome!