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The sum of two square integrable function is itself square integrable but sum of two normalized functions is not generally normalized. Why?

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    This is like asking "$1$ is finite and $1$ is finite and $1+1=2$ is finite, but why is $2\neq 1$?" Square-integrability is a finiteness condition, while normalization is an additional specification. – user580918 Aug 09 '21 at 04:16
  • Hi Mahavir, I recommend that you edit your question to include what work you've done, what effort you've made, to find an answer to your question. As written, your question is likely to be closed. – Alfred Centauri Aug 09 '21 at 13:38
  • Thank you sir for your support and useful suggestion. –  Aug 09 '21 at 14:03
  • Start with $f$ that is normalized. Then $f+f$ is not normalized, and $f-f$ is not normalized. – Disintegrating By Parts Aug 25 '21 at 03:37

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The simplest example is from geometric unit vectors. One can add together unit vectors such that the resultant vector does not have a length of $1$. The same idea holds true for square integrable functions.

BioPhysicist
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