Exercise 6.1.27

The third and the fourth condition of inner product naturally hold since

and

if $x\ne 0$. Now we prove the first two condition as Hints.

1.

Consider that

$$\parallel x+2y{\text{∥}}^2+\parallel x{\text{∥}}^2=2\parallel x+y{\text{∥}}^2+2\parallel y{\text{∥}}^2$$

and

$$\parallel x-2y{\text{∥}}^2+\parallel x{\text{∥}}^2=2\parallel x-y{\text{∥}}^2+2\parallel y{\text{∥}}^2$$

by the parallelogram law. By Substracting these two equalities we have

$$\parallel x+2y{\text{∥}}^2-\parallel x-2y{\text{∥}}^2=2\parallel x+y{\text{∥}}^2-2\parallel x-y{\text{∥}}^2.$$

And so we have

$$⟨x,2y⟩=\frac{1}{4}[\parallel x+2y{\text{∥}}^2-\parallel x-2y{\text{∥}}^2]$$

$$=\frac{1}{4}[2\parallel x+y{\text{∥}}^2-2\parallel x-y{\text{∥}}^2]=2⟨x,y⟩.$$

2.

By the previous argument, we direct prove that

$$⟨x+u,2y⟩=2⟨x,y⟩+2⟨u,y⟩.$$

Similarly begin by

$$\parallel x+u+2y{\text{∥}}^2+\parallel x-u{\text{∥}}^2=2\parallel x+y{\text{∥}}^2+2\parallel u+y{\text{∥}}^2$$

and

$$\parallel x+u-2y{\text{∥}}^2+\parallel x-u{\text{∥}}^2=2\parallel x-y{\text{∥}}^2+2\parallel u-y{\text{∥}}^2$$

by the parallelograom law. By substracting these two equalities we have

$$\parallel x+u+2y{\text{∥}}^2-\parallel x+u-2y{\text{∥}}^2=2\parallel x+y{\text{∥}}^2-2\parallel x-y{\text{∥}}^2+2\parallel u+y{\text{∥}}^2-2\parallel u-y{\text{∥}}^2.$$

And so we have

$$⟨x+u,2y⟩=\frac{1}{4}[\parallel x+u+2y{\text{∥}}^2-\parallel x+u-2y{\text{∥}}^2]$$

$$=\frac{1}{4}[2\parallel x+y{\text{∥}}^2-2\parallel x-y{\text{∥}}^2+2\parallel u+y{\text{∥}}^2-2\parallel u-y{\text{∥}}^2]$$

$$=2⟨x,y⟩+2⟨u,y⟩.$$

3.

Since $n$ is a positive integer, we have

$$⟨\mathit{nx},y⟩=⟨(n-1)x,y⟩+⟨x,y⟩$$

$$=⟨(n-2)x,y⟩+2⟨x,y⟩=\cdots =n⟨x,y⟩$$

by the previous argument inductively.

4.

Since $m$ is a positive integer, we have

$$⟨x,y⟩=⟨m(\frac{1}{m}x),y⟩=m⟨\frac{1}{m}x,y⟩$$

by the previous argument.

5.

Let $r=\frac{p}{q}$ for some positive integers $p$ and $q$ if $r$ is positive. In this case we have

$$⟨\mathit{rx},y⟩=⟨p(\frac{1}{q}x),y⟩=p⟨\frac{1}{q}x,y⟩$$

$$=\frac{p}{q}⟨x,y⟩=r⟨x,y⟩.$$

If $r$ is zero, then it’s natural that

$$⟨0,y⟩=\frac{1}{4}[\parallel y{\text{∥}}^2-\parallel y{\text{∥}}^2]=0=0⟨x,y⟩.$$

Now if $r$ is negative, then we also have

$$⟨\mathit{rx},y⟩=⟨(-r)(-x),y⟩=-r⟨-x,y⟩.$$

But we also have

$$⟨-x,y⟩=-⟨x,y⟩$$

since

$$⟨-x,y⟩+⟨x,y⟩$$

$$=\frac{1}{4}[\parallel -x+y{\text{∥}}^2-\parallel -x-y{\text{∥}}^2+\parallel x+y{\text{∥}}^2-\parallel x-y{\text{∥}}^2]=0.$$

So we know that even when $r$ is negative we have

$$⟨\mathit{rx},y⟩=-r⟨-x,y⟩=r⟨x,y⟩.$$

6.

Now we have the distribution law. Also, the position of the two component can be interchanged without change the value of the defined function. Finally we observe that

$$⟨x,x⟩=\frac{1}{4}[\parallel x+x{\text{∥}}^2-\parallel 0{\text{∥}}^2]=\parallel x{\text{∥}}^2$$

for all $x\in V$. So now we have

$$\parallel x{\text{∥}}^2+2⟨x,y⟩+\parallel y{\text{∥}}^2=\parallel x+y{\text{∥}}^2$$

$$\le {(\parallel x{\text{∥}}^{\text{+}}\parallel y\parallel )}^2=\parallel x{\text{∥}}^2+2\parallel x\parallel \cdot \parallel y\parallel +\parallel y{\text{∥}}^2$$

by the triangle inequality. Similarly we also have

$$\parallel x{\text{∥}}^2-2⟨x,y⟩+\parallel y{\text{∥}}^2=\parallel x-y{\text{∥}}^2$$

$$\le {(\parallel x{\text{∥}}^{\text{+}}\parallel -y\parallel )}^2=\parallel x{\text{∥}}^2+2\parallel x\parallel \cdot \parallel y\parallel +\parallel y{\text{∥}}^2.$$

Thus we get the desired result

$$\left|⟨x,y⟩\right|\le \parallel x\parallel \cdot \parallel y\parallel .$$

7.

Since

$$(c-r)⟨x,y⟩=c⟨x,y⟩-r⟨x,y⟩$$

and

$$⟨(c-r)x,y⟩=⟨\mathit{cx}-\mathit{rx},y⟩=⟨\mathit{cx},y⟩-r⟨x,y⟩,$$

the first equality holds. And by the previous argument we have

$$-|c-r|\parallel x\parallel \parallel y\parallel \le (c-r)⟨x,y⟩,⟨(c-r)x,y⟩\le |c-r|\parallel x\parallel \parallel y\parallel$$

and so we get the final inequality.

8.

For every real number $c$, we could find a rational number such that $|c-r|$ is small enough¹ . So by the previous argument, we have

$$⟨\mathit{cx},y⟩=c⟨x,y⟩$$

for all real number $c$.