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2.9 HW 9

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2.9.1 Section 61, Problem 2

Figure 2.86:Problem statement

Solution

Let $\psi _{2}=f+A\psi _{1}$ such that $\left \langle \psi _{2},\psi _{1}\right \rangle =0$. Hence

$$\begin{align*} \left \langle f+A\psi _{1},\psi _{1}\right \rangle & =0\\ \left \langle f,\psi _{1}\right \rangle +\left \langle A\psi _{1},\psi _{1}\right \rangle & =0\\ \left \langle f,\psi _{1}\right \rangle +A\left \langle \psi _{1},\psi _{1}\right \rangle & =0\\ \left \langle f,\psi _{1}\right \rangle +A\left \Vert \psi _{1}\right \Vert ^{2} & =0\\ A & =-\frac{\left \langle f,\psi _{1}\right \rangle }{\left \Vert \psi _{1}\right \Vert ^{2}} \end{align*}$$

Therefore, since $\psi _{2}=f+A\psi _{1}$ then

$$\psi _{2}=f-\frac{\left \langle f,\psi _{1}\right \rangle }{\left \Vert \psi _{1}\right \Vert ^{2}}\psi _{1}$$ Geometrically, the term $\frac{\left \langle \psi _{1},f\right \rangle }{\left \Vert \psi _{1}\right \Vert ^{2}}\psi _{1}$ represents the projection of $f$ on $\psi _{1}$. The term $\frac{\psi _{1}}{\left \Vert \psi _{1}\right \Vert }$ makes a unit vector in the direction of $\psi _{1}$ and the term $\frac{\left \langle f,\psi _{1}\right \rangle }{\left \Vert \psi _{1}\right \Vert }$ is the magnitude of projection $\left \Vert \psi _{1}\right \Vert \cos \left ( \theta \right )$ where $\theta$ is the inner angle between $f,\psi _{1}$. The result of $-\frac{\left \langle f,\psi _{1}\right \rangle }{\left \Vert \psi _{1}\right \Vert ^{2}}\psi _{1}$ is a vector in the opposite direction of $\psi _{1}$. Adding this to $f$ gives $\psi _{2}$ which is now orthogonal to $f$. This process is called Gram Schmidt.

2.9.2 Section 61, Problem 3

Figure 2.87:Problem statement

Solution

Let

$$\begin{align*} f & =\cos nx+\sin nx\\ \psi _{1} & =\cos nx \end{align*}$$

Then by Gram Schmidt process from problem 2 we know that

$$\psi _{2}=f-\frac{\left \langle f,\psi _{1}\right \rangle }{\left \Vert \psi _{1}\right \Vert ^{2}}\psi _{1}$$ Hence $$\begin{align*} \psi _{2} & =\left ( \cos nx+\sin nx\right ) -\frac{\int _{-\pi }^{\pi }\left ( \cos nx+\sin nx\right ) \cos nxdx}{\int _{-\pi }^{\pi }\cos ^{2}\left ( nx\right ) dx}\cos nx\\ & =\left ( \cos nx+\sin nx\right ) -\frac{\int _{-\pi }^{\pi }\cos nx\cos nxdx+\int _{-\pi }^{\pi }\sin nx\cos nxdx}{\pi }\cos nx \end{align*}$$

But $\int _{-\pi }^{\pi }\cos nx\cos nxdx=\int _{-\pi }^{\pi }\cos ^{2}nxdx=\pi$ and $\int _{-\pi }^{\pi }\sin nx\cos nxdx=0$ since these are orthogonal. Hence the above simplifies to

$$\begin{align*} \psi _{2} & =\left ( \cos nx+\sin nx\right ) -\cos nx\\ & =\sin nx \end{align*}$$

2.9.3 Section 63, Problem 3

Figure 2.88:Problem statement

Solution

The Fourier coefficients of $f-g$ are given by $\left \langle f-g,\phi _{n}\right \rangle$ by definition. But due to linearity of inner product, this can be written as

$$\left \langle f-g,\phi _{n}\right \rangle =\left \langle f,\phi _{n}\right \rangle -\left \langle g,\phi _{n}\right \rangle$$ But $\left \langle f,\phi _{n}\right \rangle$ are the Fourier coefficients of $f$ and $\left \langle g,\phi _{n}\right \rangle$ are the Fourier coefficients of $g$, and we are told these are the same. Therefore $$\left \langle f-g,\phi _{n}\right \rangle =0$$ Which implies that $\left \Vert f-g\right \Vert =0$. Using part(b) in problem 4, section 61, which says that if $\left \Vert f\right \Vert =0$ then $f\left ( x\right ) =0$ except at possibly finite number of points in the interval, then applying this to $\left \Vert f-g\right \Vert =0$ leads to $$f-g=0$$ Which implies $f=g$ which is what required to show.

2.9.4 Section 63, Problem 4

Figure 2.89:Problem description

solution

Part (a)

Let the generalized Fourier series of $f\left ( x\right )$ be

$$f\left ( x\right ) =\sum _{n=1}^{\infty }\left \langle f\left ( x\right ) ,\phi _{n}\right \rangle \phi _{n}$$ Let the sum the above converges uniformly to be $s\left ( x\right )$. Therefore we have, per problem statement the following equality $$\sum _{n=1}^{\infty }\left \langle f\left ( x\right ) ,\phi _{n}\right \rangle \phi _{n}=s\left ( x\right )$$ Taking the inner product of both sides with respect to $\phi _{m}$ gives $$\begin{align*} \int _{a}^{b}\left ( \sum _{n=1}^{\infty }\left \langle f\left ( x\right ) ,\phi _{n}\right \rangle \phi _{n}\right ) \phi _{m}dx & =\int _{a}^{b}s\left ( x\right ) \phi _{m}dx\\ & =\left \langle s\left ( x\right ) ,\phi _{m}\right \rangle \end{align*}$$

Since the sum converges uniformly, then we are allowed to integrate the left side term by term while keeping the equality with the right side. Hence moving the integration inside the sum gives

$$\sum _{n=1}^{\infty }\left \langle f\left ( x\right ) ,\phi _{n}\right \rangle \int _{a}^{b}\phi _{n}\phi _{m}dx=\left \langle s\left ( x\right ) ,\phi _{m}\right \rangle$$ But due to orthogonality of $\phi _{n}$ and $\phi _{m}$ and since they are normalized, then $\int _{a}^{b}\phi _{n}\phi _{m}dx=\left \langle \phi _{n},\phi _{m}\right \rangle =1$ if $n=m$ and zero otherwise. Hence the above simplifies to $$\left \langle f\left ( x\right ) ,\phi _{m}\right \rangle =\left \langle s\left ( x\right ) ,\phi _{m}\right \rangle$$ And since the above is valid for any arbitrary $m=1\cdots \infty$, then it shows that $f\left ( x\right )$ and $s\left ( x\right )$ have the same generalized Fourier coefficients.

Part (b)

From part (a), we found

$$\left \langle f,\phi _{n}\right \rangle =\left \langle s,\phi _{n}\right \rangle$$ By linearity of inner product, the above is the same as $$\begin{align*} \left \langle f,\phi _{n}\right \rangle -\left \langle s,\phi _{n}\right \rangle & =0\\ \left \langle f-s,\phi _{n}\right \rangle & =0 \end{align*}$$

But from problem 3, we know that $\left \langle f-s,\phi _{n}\right \rangle =0$ implies $\left \Vert f-s\right \Vert =0$.

Next, using part(b) in problem 4, section 61, which says that if $\left \Vert f\right \Vert =0$ then $f\left ( x\right ) =0$ except at possibly finite number of points in the interval, then applying this to our case here that $\left \Vert f-s\right \Vert =0$ leads to

$$\begin{align*} f-s & =0\\ f & =s \end{align*}$$

Which is the result required to show.

2.9.5 Section 66, Problem 4

Figure 2.90:Problem description

solution

Part (a)

We need to find

$$\begin{align*} & \left \langle \phi _{0},\phi _{2n}\right \rangle \\ & \left \langle \phi _{0},\phi _{2n-1}\right \rangle \\ & \left \langle \phi _{2n},\phi _{2m}\right \rangle \\ & \left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle \\ & \left \langle \phi _{2m-1},\phi _{2n}\right \rangle \end{align*}$$

And also show that

$$\begin{align*} & \left \langle \phi _{0},\phi _{0}\right \rangle =\left \Vert \phi _{0}\right \Vert ^{2}=1\\ & \left \langle \phi _{2n},\phi _{2n}\right \rangle =\left \Vert \phi _{2n}\right \Vert ^{2}=1\\ & \left \langle \phi _{2n-1},\phi _{2n-1}\right \rangle =\left \Vert \phi _{2n-1}\right \Vert ^{2}=1 \end{align*}$$

$\left \langle \phi _{0},\phi _{2n}\right \rangle$

$$\begin{align*} \left \langle \phi _{0},\phi _{2n}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{2c}}\frac{1}{\sqrt{c}}\cos \left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c\sqrt{2}}\left [ \frac{\sin \left ( \frac{n\pi }{c}x\right ) }{\frac{n\pi }{c}}\right ] _{-c}^{c}\\ & =\frac{c}{n\pi c\sqrt{2}}\left [ \sin \left ( \frac{n\pi }{c}x\right ) \right ] _{-c}^{c}\\ & =\frac{1}{n\pi \sqrt{2}}\left [ \sin \left ( n\pi \right ) +\sin \left ( n\pi \right ) \right ] \\ & =0 \end{align*}$$

Since $n$ is integer.

$\left \langle \phi _{0},\phi _{2n-1}\right \rangle$

$$\begin{align*} \left \langle \phi _{0},\phi _{2n-1}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{2c}}\frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c\sqrt{2}}\left [ \frac{-\cos \left ( \frac{n\pi }{c}x\right ) }{\frac{n\pi }{c}}\right ] _{-c}^{c}\\ & =\frac{-c}{n\pi c\sqrt{2}}\left [ \cos \left ( \frac{n\pi }{c}x\right ) \right ] _{-c}^{c}\\ & =\frac{-1}{n\pi \sqrt{2}}\left [ \cos \left ( n\pi \right ) -\cos \left ( n\pi \right ) \right ] \\ & =0 \end{align*}$$

$\left \langle \phi _{2n},\phi _{2m}\right \rangle$

$$\begin{align*} \left \langle \phi _{2n},\phi _{2m}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi }{c}x\right ) \frac{1}{\sqrt{c}}\sin \left ( \frac{m\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\sin \left ( \frac{n\pi }{c}x\right ) \sin \left ( \frac{m\pi }{c}x\right ) dx \end{align*}$$

Let $\frac{c}{\pi }s=x$, then $dx=\frac{c}{\pi }ds$. When $x=-c$ then $s=-\pi$ and when $x=c$ then $s=\pi$ and the above becomes

$$\begin{align*} \left \langle \phi _{2n},\phi _{2m}\right \rangle & =\frac{1}{c}\int _{-\pi }^{\pi }\sin \left ( ns\right ) \sin \left ( ms\right ) \frac{c}{\pi }ds\\ & =\frac{1}{\pi }\int _{-\pi }^{\pi }\sin \left ( ns\right ) \sin \left ( ms\right ) ds \end{align*}$$

Since the integrand is even, then

$$\left \langle \phi _{2n},\phi _{2m}\right \rangle =\frac{2}{\pi }\int _{0}^{\pi }\sin \left ( ns\right ) \sin \left ( ms\right ) ds$$ From equation (1), page 192 we see that

$$\left \langle \phi _{2n},\phi _{2m}\right \rangle =0$$ Since $n,m$ are different.

$\left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle$

$$\begin{align*} \left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{c}}\cos \left ( \frac{n\pi }{c}x\right ) \frac{1}{\sqrt{c}}\cos \left ( \frac{m\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\cos \left ( \frac{n\pi }{c}x\right ) \cos \left ( \frac{m\pi }{c}x\right ) dx \end{align*}$$

Let $\frac{c}{\pi }s=x$, then $dx=\frac{c}{\pi }ds$. When $x=-c$ then $s=-\pi$ and when $x=c$ then $s=\pi$ and the above becomes

$$\begin{align*} \left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle & =\frac{1}{c}\int _{-\pi }^{\pi }\cos \left ( ns\right ) \cos \left ( ms\right ) \frac{c}{\pi }ds\\ & =\frac{1}{\pi }\int _{-\pi }^{\pi }\cos \left ( ns\right ) \cos \left ( ms\right ) ds \end{align*}$$

Since the integrand is even, then

$$\left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle =\frac{2}{\pi }\int _{0}^{\pi }\cos \left ( ns\right ) \cos \left ( ms\right ) ds$$ From equation (4), page 192 we see that $$\left \langle \phi _{2n-1},\phi _{2m-1}\right \rangle =0$$ Since $n,m$ are different.

$\left \langle \phi _{2m-1},\phi _{2n}\right \rangle$

$$\begin{align*} \left \langle \phi _{2m-1},\phi _{2n}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{c}}\cos \left ( \frac{m\pi }{c}x\right ) \frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\cos \left ( \frac{m\pi }{c}x\right ) \sin \left ( \frac{n\pi }{c}x\right ) dx \end{align*}$$

Let $\frac{c}{\pi }s=x$, then $dx=\frac{c}{\pi }ds$. When $x=-c$ then $s=-\pi$ and when $x=c$ then $s=\pi$ and the above becomes

$$\begin{align*} \left \langle \phi _{2m-1},\phi _{2n}\right \rangle & =\frac{1}{c}\int _{-\pi }^{\pi }\cos \left ( ms\right ) \sin \left ( ns\right ) \frac{c}{\pi }ds\\ & =\frac{1}{\pi }\int _{-\pi }^{\pi }\cos \left ( ms\right ) \sin \left ( ns\right ) ds \end{align*}$$

Using $\cos \left ( ms\right ) \sin \left ( ns\right ) =\frac{1}{2}\left ( \cos \left ( s\left ( m+n\right ) \right ) +\cos \left ( s\left ( m-n\right ) \right ) \right )$. Hence the above becomes

$$\left \langle \phi _{2m-1},\phi _{2n}\right \rangle =\frac{1}{2\pi }\left ( \int _{-\pi }^{\pi }\cos \left ( s\left ( m+n\right ) \right ) ds+\int _{-\pi }^{\pi }\cos \left ( s\left ( m-n\right ) \right ) ds\right )$$ Since the integration is over one full period, then each is zero. Hence $$\left \langle \phi _{2m-1},\phi _{2n}\right \rangle =0$$ $\left \langle \phi _{0},\phi _{0}\right \rangle$ $$\begin{align*} \left \langle \phi _{0},\phi _{0}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{2c}}\frac{1}{\sqrt{2c}}dx\\ \left \Vert \phi _{0}\right \Vert ^{2} & =\frac{1}{2c}\int _{-c}^{c}dx\\ & =1 \end{align*}$$

Hence $\left \Vert \phi _{0}\right \Vert =1\,$.

$\left \langle \phi _{2n},\phi _{2n}\right \rangle$

$$\begin{align*} \left \langle \phi _{2n},\phi _{2n}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi }{c}x\right ) \frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\sin ^{2}\left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\frac{1}{2}-\frac{1}{2}\cos \left ( 2\frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{2c}\left ( \int _{-c}^{c}dx-\int _{-c}^{c}\cos \left ( 2\frac{n\pi }{c}x\right ) dx\right ) \\ & =\frac{1}{2c}\left ( 2c-\left [ \frac{\sin \left ( 2\frac{n\pi }{c}x\right ) }{2\frac{n\pi }{c}}\right ] _{-c}^{c}\right ) \\ & =\frac{1}{2c}\left ( 2c-\frac{c}{2n\pi }\left [ \sin \left ( 2\frac{n\pi }{c}x\right ) \right ] _{-c}^{c}\right ) \\ & =\frac{1}{2c}\left ( 2c\right ) \\ & =1 \end{align*}$$

Hence $\left \Vert \phi _{2n}\right \Vert =1\,$.

$\left \langle \phi _{2n-1},\phi _{2n-1}\right \rangle$

$$\begin{align*} \left \langle \phi _{2n-1},\phi _{2n-1}\right \rangle & =\int _{-c}^{c}\frac{1}{\sqrt{c}}\cos \left ( \frac{n\pi }{c}x\right ) \frac{1}{\sqrt{c}}\cos \left ( \frac{n\pi }{c}x\right ) dx\\ \left \Vert \phi _{2n-1}\right \Vert ^{2} & =\frac{1}{c}\int _{-c}^{c}\cos ^{2}\left ( \frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{c}\int _{-c}^{c}\frac{1}{2}+\frac{1}{2}\sin \left ( 2\frac{n\pi }{c}x\right ) dx\\ & =\frac{1}{2c}\left ( \int _{-c}^{c}dx+\int _{-c}^{c}\sin \left ( 2\frac{n\pi }{c}x\right ) dx\right ) \\ & =\frac{1}{2c}\left ( 2c-\left [ \frac{\cos \left ( 2\frac{n\pi }{c}x\right ) }{2\frac{n\pi }{c}}\right ] _{-c}^{c}\right ) \\ & =\frac{1}{2c}\left ( 2c-\frac{c}{2n\pi }\left [ \cos \left ( 2\frac{n\pi }{c}x\right ) \right ] _{-c}^{c}\right ) \\ & =\frac{1}{2c}\left ( 2c-\frac{c}{2n\pi }\left [ \cos \left ( 2n\pi \right ) -\cos \left ( 2n\pi \right ) \right ] \right ) \\ & =\frac{1}{2c}2c\\ & =1 \end{align*}$$

Hence $\left \Vert \phi _{2n-1}\right \Vert =1\,$.

Part (b)

$$\begin{align*} \phi _{0}\left ( x\right ) & =\frac{1}{\sqrt{2c}}\\ \phi _{2n-1}\left ( x\right ) & =\frac{1}{\sqrt{c}}\cos \left ( \frac{n\pi x}{c}\right ) \\ \phi _{2n}\left ( x\right ) & =\frac{1}{\sqrt{c}}\sin \left ( \frac{n\pi x}{c}\right ) \end{align*}$$

On $-c<x<c$. The generalized Fourier series for $f\left ( x\right )$ in $C_{p}\left ( -c,c\right )$ is

$$\sum _{n=0}^{\infty }c_{n}\phi _{n}\left ( x\right ) =c_{0}\phi _{0}\left ( x\right ) +\sum _{n=1}^{\infty }\left ( c_{2n-1}\phi _{2n-1}\left ( x\right ) +c_{2n}\phi _{2n}\left ( x\right ) \right )$$ That is $$\begin{equation} f\left ( x\right ) \sim c_{0}\frac{1}{\sqrt{2c}}+\sum _{n=1}^{\infty }\left ( \frac{c_{2n-1}}{\sqrt{c}}\cos \left ( \frac{n\pi x}{c}\right ) +\frac{c_{2n}}{\sqrt{c}}\sin \left ( \frac{n\pi x}{c}\right ) \right ) \tag{1} \end{equation}$$ Where $$c_{0}=\left \langle f,\phi _{0}\left ( x\right ) \right \rangle =\frac{1}{\sqrt{2c}}\int _{-c}^{c}f\left ( x\right ) dx$$ And $$\begin{align*} c_{2n-1} & =\left \langle f,\phi _{2n-1}\left ( x\right ) \right \rangle =\frac{1}{\sqrt{c}}\int _{-c}^{c}f\left ( x\right ) \cos \left ( \frac{n\pi x}{c}\right ) dx\qquad n=1,2,\cdots \\ c_{2n} & =\left \langle f,\phi _{2n}\left ( x\right ) \right \rangle =\frac{1}{\sqrt{c}}\int _{-c}^{c}f\left ( x\right ) \sin \left ( \frac{n\pi x}{c}\right ) dx\qquad n=1,2,\cdots \end{align*}$$

If we write

$$a_{0}=2\frac{c_{0}}{\sqrt{2c}},a_{n}=\frac{c_{2n-1}}{\sqrt{c}},b_{n}=\frac{c_{2n}}{\sqrt{c}}\qquad n=1,2,\cdots$$ Then (1) becomes $$f\left ( x\right ) \sim \frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( \frac{n\pi x}{c}\right ) +b_{n}\sin \left ( \frac{n\pi x}{c}\right )$$ Where $$\begin{align*} a_{n} & =\frac{1}{c}\int _{-c}^{c}f\left ( x\right ) \cos \left ( \frac{n\pi x}{c}\right ) dx\qquad n=1,2,\cdots \\ b_{n} & =\frac{1}{c}\int _{-c}^{c}f\left ( x\right ) \sin \left ( \frac{n\pi x}{c}\right ) dx\qquad n=1,2,\cdots \end{align*}$$

This is the ordinary Fourier series on $-c<x<c$.

Part (c)

From (1) section 65

$$\begin{equation} \sum _{n=0}^{N}c_{n}^{2}\leq \left \Vert f\right \Vert ^{2} \tag{1} \end{equation}$$ But from part (b) we found that $$a_{0}=2\frac{c_{0}}{\sqrt{2c}},a_{n}=\frac{c_{2n-1}}{\sqrt{c}},b_{n}=\frac{c_{2n}}{\sqrt{c}}\qquad n=1,2,\cdots$$ Hence $$\begin{align*} c_{0} & =\frac{a_{0}}{2}\sqrt{2c}\\ c_{2n-1} & =a_{n}\sqrt{c}\\ c_{2n} & =b_{n}\sqrt{c} \end{align*}$$

Substituting the above into (1) gives

$$\begin{align*} c_{0}^{2}+\sum _{n=1}^{N}c_{2n-1}^{2}+\sum _{n=1}^{N}c_{2n}^{2} & \leq \left \Vert f\right \Vert ^{2}\\ \left ( \frac{a_{0}}{2}\sqrt{2c}\right ) ^{2}+\sum _{n=1}^{N}\left ( a_{n}\sqrt{c}\right ) ^{2}+\sum _{n=1}^{N}\left ( b_{n}\sqrt{c}\right ) ^{2} & \leq \int \left [ f\left ( x\right ) \right ] ^{2}dx\\ \left ( \frac{a_{0}^{2}}{4}2c\right ) +\sum _{n=1}^{N}a_{n}^{2}c+\sum _{n=1}^{N}b_{n}^{2}c & \leq \int \left [ f\left ( x\right ) \right ] ^{2}dx\\ \frac{a_{0}^{2}}{2}+\sum _{n=1}^{N}\left ( a_{n}^{2}+b_{n}^{2}\right ) & \leq \frac{1}{c}\int \left [ f\left ( x\right ) \right ] ^{2}dx \end{align*}$$

2.9.6 Section 66, Problem 5

Figure 2.91:Problem description

solution

The function $S_{N}\left ( x\right )$ is almost $1$ everywhere as can be seen from this diagram

Figure 2.92:Showing the function $S_N(x)$ and $f(x)$

And the problem is asking us to show that $S_{N}\left ( x\right ) \rightarrow f\left ( x\right )$ in the mean.  This means we need to show the following is true

$$\lim _{N\rightarrow \infty }\left \Vert S_{N}\left ( x\right ) -f\left ( x\right ) \right \Vert =0$$

Except at possibly finite number of points $x$. But this is the case here. Looking at $S_{N}\left ( x\right )$ we see it is equal to $f\left ( x\right ) =1$ everywhere except at the points $x=1,\frac{1}{2},\frac{1}{3},\cdots$ and compared to all the points between $0$ and $1$, then $S_{N}\left ( x\right ) =f\left ( x\right ) =1$ almost everywhere. Even though as $N\rightarrow \infty$ the number of points where $S_{N}\left ( x\right ) \neq 1$ increases, it is still finitely many compared to the number of points where $S_{N}\left ( x\right ) =f\left ( x\right ) =1$.

To answer the second part:  Since $S_{N}\left ( x\right ) =0$ at any $x$ value which can written as $\frac{1}{p}$ where $p$ is an integer (this by definition given), then $S_{N}\left ( \frac{1}{p}\right ) =0$. Then it clearly follows that $\lim _{N\rightarrow \infty }S_{N}\left ( \frac{1}{p}\right ) =0$.