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I would like to show that

If $f\in C^0([a,b]),p\geq 1$ then $\int^b_a|f(x)|^pdx=0$ implies $f(x)=0$ $\forall x \in [a,b].$

This is how I argue:

Since $|f(x)|\geq 0$ and $f$ is continuous then the only way the integral can equal $0$ is if $f(x)=0$ for all $x \in [a,b]$

Would this be wrong or not rigorous enough?

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    $\begingroup$ No, not rigorous. Where did you use continuity? $\endgroup$ – Kavi Rama Murthy May 19 '20 at 12:19
  • $\begingroup$ @KaviRamaMurthy Do I need to use it? It is said in the definition of the function. But could you explain what exactly i'm missing? $\endgroup$ – user634512 May 19 '20 at 12:20
  • $\begingroup$ in your proof what happens when p is 0 $\endgroup$ – Monocerotis May 19 '20 at 12:21
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    $\begingroup$ If $f(x)=0$ for $x \neq 0$ and $1$ for $x=0$ then $\int |f|^{p}$ exists and equals $0$ but the function is not $0$. Continuity is very important. $\endgroup$ – Kavi Rama Murthy May 19 '20 at 12:23
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    $\begingroup$ Why is it that the only way the integral can be zero is when $f\equiv 0$? $\endgroup$ – Sam May 19 '20 at 12:25
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As mentioned in the comments your argument was not rigorous as you didn't justify your statement: "the only way the integral can equal $0$ is if $f(x)=0$ for all $x \in [a,b]$."

Proof: Suppose there exists $x_0 \in [a,b]$ such that $f(x_0) \neq 0.$ Then $\epsilon=|f(x_0)|^p>0.$ Continuity of $f$ implies there exists $\delta>0$ such that \begin{align*}x\in[a,b],|x-x_0|<\delta &\implies ||f|^p(x)-|f|^p(x_0)|<\frac{\epsilon}{2}\\&\implies ||f|^p(x)-\epsilon|<\frac{\epsilon}{2}\\&\implies|f(x)|^p>\frac{\epsilon}{2}.\end{align*}

Therefore $$\int_a^b |f(x)|^p\,dx\geq \int_{x_0-\delta}^{x_0+\delta}|f(x)|^p\,dx>\epsilon\delta>0$$ which is a contradiction.

(Note: You can always choose $\delta>0$ small enough so that $(x_0-\delta,x_0+\delta)\subseteq [a,b].$)

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    $\begingroup$ just a small doubt from a student. Here $||f|^p(x)-|f|^p(x_0)|<\frac{\epsilon}{2}$ why have you specifically chosen $\frac{\epsilon}{2}$ please... $\endgroup$ – Monocerotis May 19 '20 at 14:12
  • $\begingroup$ @Monocerotis Oh you could choose anything less than $\epsilon,$ just to obtain the final term is $>0.$ $\endgroup$ – Sahiba Arora May 19 '20 at 16:32
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    $\begingroup$ got it.thanks a lot! $\endgroup$ – Monocerotis May 20 '20 at 2:32
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    $\begingroup$ Many compliments for your career. $\endgroup$ – Sebastiano May 20 '20 at 22:46

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