A function $f(x) = 4x + 1$ and $g(x) = \sqrt{3-x}$ are vectors inside inner product space $C [0,2]$, with the inner product inside it defined as: $(f, g) = \int_0^2{f(x)g(x) dx}$.
So find: Norm from the functions $f$ and $g$, distance between $f$ and $g$ inside the inner product, and a function which is orthogonal to $f$.
Can you help me with a little bit of explanation? Thank you so much!
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Norm: $\|f\|=\sqrt{(f,f)}=\sqrt{\int_0^2 f^2(x)dx}$. Distance between $f$ and $g$ is $\|f-g\|$. The function $h$ is orthogonal to $f$ iff $(f,h)=0$. Try to choose $h$ e.g. between constant functions.
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The norm of $f$ is $\sqrt{(f,f)}=\sqrt{\int_0^2 f^2(x) dx}$.
A function $h$ orthogonal to $f$ satisfies $(f,h)=0$ (obviously $h=0$ is orthogonal to anything). For any function $g$ , you can easily see that $g - \frac{(g,f)}{(f,f)} f$ is orthogonal to $f$ (this is a special case of the Gram-Schmidt process; if $g = \alpha f$ for some scalar $\alpha$, then $g - \frac{(g,f)}{(f,f)} f=0$).
Not sure what distance of between f and g inside the inner product means, but it likely means the norm of $f-g$. Note that $(f-g,f-g) = (f,f) +(g,g) - 2 (f,g)$ by bi-linearity of the inner product. So, you only need to compute $(f,g)$ to get the norm of $f-g$ assuming you've computed the norm of $f$ and $g$.
The norm of an element $x$ of an inner product space with inner product $(\cdot,\cdot)$ is defined by $\|x\|=\sqrt{(x,x)}$. The distance between two elements $x,y$ in the inner product space is defined as the norm of their difference, i.e. $\|x-y\|$.
So in your problem, to find the norm of $f$ you compute $$\|f\|=(f,f)^{1/2}=\left(\int_0^2\left(f(x)\right)^2dx\right)^{1/2}.$$
Similarly, the distance between $f$ and $g$ is $$\|f-g\|=\left(\int_0^2\left(f(x)-g(x)\right)^2dx\right)^{1/2}.$$
To find a function orthogonal to $f$ is to find a function $h\in C([0,2])$ such that $(f,h)=0$. My hint would be to stick with polynomials, perhaps $h(x)=ax+b$ and solve for $a$ and $b$.