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The dual space of a normed linear space $V$ is the space of all linear bounded functional on $V$:

$$ V^*:=\{f:V\to R\mid\text{$f$ is linear and bounded}\} $$ The norm of $V^*$ is defined as: $$ \|f\|=\sup_{\|u\| \leq 1} |f(u)| \tag{1} $$

Can you explain to me why (1) is equivalence to the definition: $$ \|f\|=\sup \frac{|f(u)|}{\|u\|} $$

I tried to work this out myself but I could only show the equivalence when the $\|u\| \leq 1$ in (1) is changed into $\|u\|=1$.

Michael Hardy
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A Slow Learner
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2 Answers2

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By linearity, $\frac{|f(u)|}{\|u\|} = |f(\frac{u}{\|u\|})|$ so that's one direction, that is (2) (the 2nd def) is $\leq (1)$. The other direction follows from the fact that if $\|u\|\leq 1$, then $|f(u)| \leq \frac{|f(u)|}{\|u\|}$

Michael Hardy
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ureui
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  • Can you elaborate more on why (2) < (1)? – A Slow Learner Sep 15 '20 at 04:27
  • Let $v = \frac{u}{||u||}$. Then $||v|| = 1$ and you're taking a larger sup, indeed for $||v||\leq 1$. – ureui Sep 15 '20 at 12:29
  • I completely understand the proof now. However, I am still having trouble seeing the "image" of the proof. In fact, I have this same problem whenever a proof on equality of two quantities involving proving each side is smaller or equal (or larger or equal) to the other. Do you think you can explain a bit more to me how this particular proof works? – A Slow Learner Sep 16 '20 at 16:31
  • I dont know what you mean, but if $a\leq b\leq a$, then $a=b$. – ureui Sep 17 '20 at 11:27
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When $\|u\|\lt 1$, define $v=\frac{u}{\|u\|}$ Let $c=\frac{1}{\|u\|}$. Since the operator is linear $f(u)=cf(v)$, and $\frac{|f(u)|}{\|u\|}=\frac{c|f(v)|}{c\|v\|}$, where $\|v\|=1$.

Michael Hardy
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herb steinberg
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