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I have read in a textbook that '' if a function f(x) is continuous then $$\left\|{f}\right\|=\sqrt{\left({f(x),f(x)}\right)}=\sqrt{\int\limits_{a}^{b}{r(x)f(x)^{2}dx}}=0$$ implies that $$f(x)\equiv 0$$ (The weight function r is positive) How is it possible that a non-zero discontinuous function has zero norm?

absolutezero
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    If you change a function at a single point (or a countable number of points), you won't change its integral, hence you won't change its norm, and so a function with norm zero only has to be zero "almost everywhere" (a technical term). But a continuous function that is zero almost everywhere is actually zero everywhere. – Aaron Dec 03 '22 at 18:37
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    What is $r$ in your quote? + replace "continues" by "continuous" + Your title is not coherent with your question. – Anne Bauval Dec 03 '22 at 18:38
  • @AnneBauval r is the weight function that is norm of f is zero with respect to the weight function r(x) – absolutezero Dec 03 '22 at 18:40
  • I knew, but better add it into your post than as a comment, and with the appropriate hypothesis on $r.$ – Anne Bauval Dec 03 '22 at 18:41

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