(1) Show that any inner product space $V$ is a normed space with norm $||f||=\sqrt{\langle f, f \rangle}$, $\forall f\in V$.

Can someone please explain what kind of proof is required here? I've seen something like:

(2) If $X$ is a linear space with inner product $(.,.)$, then we can define a norm on $X$ by $||x|| = \sqrt{(x,x)}$. Thus, any inner product space is a normed linear space.

Where is the "proof" here? It seems like it is saying the same thing as $(1)$, so what is required of a proof to show the result of $(1)$?

  • $\begingroup$ Were you given the norm axioms that any norm must satisfy? $\endgroup$ – JohnD Nov 11 '14 at 21:51
  • $\begingroup$ @JohnD Yes, I was. $\endgroup$ – Caleb J Nov 11 '14 at 21:52
  • $\begingroup$ @CalebJ Do you understand what's being asked of you in the first question? $\endgroup$ – Git Gud Nov 11 '14 at 21:52
  • $\begingroup$ @CalebJ: I see you have 7 questions but no accepted answers. math.stackexchange.com/help/someone-answers $\endgroup$ – JohnD Nov 11 '14 at 23:23

Since you have the norm axioms, verify that they each hold:

  1. $\|c \mathbf{x}\|=|c|\|\mathbf{x}\|$ for all scalars $c$ and vectors $\mathbf x$

  2. $\|\mathbf{x}+\mathbf{y}\|\le\|\mathbf{x}\|+\|\mathbf{y}\|$ for all vectors $\mathbf x, \mathbf y$

  3. $\|\mathbf x\|\ge 0$ with equality if and only if $\mathbf x=\mathbf 0$.

Just do this using $\|\mathbf x\|:=\sqrt{\langle\mathbf x,\mathbf x\rangle}$ with the scalars $c$ from whatever your underlying field is (probably $\mathbb R$ or $\mathbb C$).

  • $\begingroup$ Can you please do one of these as an example? $\endgroup$ – Caleb J Nov 11 '14 at 21:57
  • $\begingroup$ Fall back on the properties of the underlying inner product: $\|cx\|=\sqrt{\langle cx,cx\rangle}=\sqrt{c^2\langle x, x\rangle}=\sqrt{c^2}\,\sqrt{\langle x,x,\rangle}=|c|\|x\|$ $\endgroup$ – JohnD Nov 11 '14 at 22:01
  • $\begingroup$ The other two are similar. The key is you will need to use your knowledge of the inner product axioms along the way for each of the three parts. $\endgroup$ – JohnD Nov 11 '14 at 22:13

The last condition to show that $|x| = \sqrt{\langle x,x \rangle}$ is a norm (the other two are quite simple) should go like this:

$$\begin{align}|x+y|^2 &= \langle x+y,x+y\rangle = |x|^2 + |y|^2 + 2\langle x,y\rangle \\ &\leq |x|^2 + |y|^2 + 2| x| \dot \ |y| = (|x|+|y|)^2\end{align} $$

The question says that from an inner product, we may define the norm of a vector $x \in E$ by taking $|x| = \sqrt{\langle x,x\rangle} $. Not every normed space $E$, however, comes from an inner product. This will only occur when the so called Parallelogram Law holds:

$$|x+y|^2 + |x-y|^2 = 2(|x|+|y|)^2$$

For example the norm $|x|' = \sum |x_i| $ in $\mathbb{R}^n$ is not yielded from an inner product because the Parallelogram Law does not hold.

If you have any question don't hesitate to ask on the comments.


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