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I have a brief question regarding the norm of the $L^2$ space defined on an interval $[a,b]$. On various websites I have seen this defined as:

$$\|f(x)\| = \int_{a}^{b} f(x)^2 dx$$

However, yesterday I posted a question in which I had to use:

$$\|f(x)\| = \sqrt{\int_{a}^{b} f(x)^2 dx}$$

to get the proper answer. I asked for clarification about this in yesterday's thread, but did not receive an explanation - probably because by the time I inquired about this, my thread was already getting old. The link to my question from yesterday is:

Gram-Schmidt Orthogonalization for subspace of $L^2$

If anyone could please clarify for me exactly why we use the norm definition with a square root in the link above, I would be extremely grateful!

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3 Answers 3

up vote 2 down vote accepted

One of the norm's axiom is semilinearity:$$\| \lambda x\|=|\lambda| \cdot\| x\|\quad \forall \lambda \in \mathbb{R} \;(\mathrm {or}\;\mathbb{C})$$

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You're absolutely right! Thanks a lot. –  Kristian Sep 25 '12 at 8:53

The $L_2$ norm is defined by your second formula: $$\|f(x)\| = \sqrt{\int_{a}^{b} f(x)^2 dx}$$ or, equivalently $$\|f(x)\|^2 = {\int_{a}^{b} f(x)^2 dx}.$$

Your first formula defines a “norm” that doesn't satisfy basic norm properties. In particular, $\|\alpha f\| \neq \alpha \|f\|$ for $\alpha\in {\mathbb R}_{\geq 0}$ (but rather $\|\alpha f\| = \alpha^2 \|f\|$).

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Thanks a lot! I must have visited a shady website, where the first definition was used! I'm really happy to have this cleared up! Appreciate it a lot. –  Kristian Sep 25 '12 at 8:52

Correct definition of $\|f(x)\| $ is not

$$\|f(x)\| = \int_{a}^{b} f(x)^2 dx$$ but

$$\|f(x)\| = \sqrt{\int_{a}^{b} f(x)^2 dx}$$

see http://en.wikipedia.org/wiki/Norm_(mathematics)#Infinite-dimensional_case

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Thanks a lot. Yeah, there must have been a typo where I found the first definition. Really appreciate your answer! –  Kristian Sep 25 '12 at 8:53

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