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Define two norms as following: $$ \left\Vert f\right\Vert _{1}={ \max_{0\leq x\leq1}\left|f\left(x\right)\right|} , \quad\text{ and }\quad \left\Vert f\right\Vert _{2}={\intop_{0}^{1}\left|f\left(x\right)\right|dx} $$

on the vector space $ C\left[0,1\right] $ (the continuous functions).

I need to prove that the two norms aren't equivalent.

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Let $f_n(x):=\max\{1-nx,0\}$, which is a continuous function for all $n$. The function $f_n$ is non-negative and reaches its maximum at $x=0$ hence $\lVert f_n\rVert_1=1$. However, the computation of the $\lVert\cdot\rVert_2$-norm leads to the computation of the area of a triangle, which is half of that of a rectangle having corners $(0,0)$ and $(1/n,1)$, hence $\lVert f_n\rVert_2=1/(2n)$.

This shows that there does not exist a constant $C$ such that for each $f\in C[0,1]$, $\lVert f\rVert_1\leqslant C\lVert f\rVert_2$.

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  • $\begingroup$ Diverge to infinity. $\endgroup$ Nov 27, 2012 at 15:43
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    $\begingroup$ @MarekKryspin Thanks for the catch. I rewrote completely this answer. $\endgroup$ Nov 9, 2023 at 10:19
  • $\begingroup$ +1. $\phantom{}$ $\endgroup$ Nov 9, 2023 at 15:14

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