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If $$\|f \| =\sup \{|f(x)|:x \in [0,1]\} $$ and also $$ \|f \|=\int^1_0 |f(x)| \, dx,$$ then for $f(x)=x$, we have $\sup \{|f(x)|:x \in [0,1]\} = 1$. But $\int^1_0 |f (x)| \, dx = \int^1_0 |x| \, dx= \frac{1}{2}$. Then how is it possible ?

Michael Hardy
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my stak
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  • They are two different norms. More precisely, one should write $|f|_\infty$ and $|f|_1$ for the two. – Quang Hoang Aug 16 '14 at 14:52
  • This is just like defining a metric space with different metrics. You can equip the real line $\mathbb{R}$ with the normal Euclidean metric, i.e., $ d(x,y) = |x - y| $, or, maybe, with the discrete metric, i.e., $d(x,y) = \begin{cases} 0 & x = y \ 1 & x \neq y \end{cases}$. With the Euclidean metric, $d(\frac{1}{2}, 1) = \frac{1}{2}$, while with the discrete metric, $d(\frac{1}{2}, 1) = 1$. – layman Aug 16 '14 at 14:57

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A vector space can be associated with different norms, all satisfying norm axioms. In your example, the first is $L^\infty$ norm and the second $L^1$ norm.

Shuchang
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