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Show that if $a>1$ then $$\|f||=\min \left\{\max\{|f(t)| : t \in [0,1]\}, a \int_{0}^{1} |f(t)|dt \right\}$$ is not a norm in $C[0,1]$.

Can someone help? My idea is to find some functions $f$ and $g$ in $C[0,1]$ such that $\|f+g\| > \|f\|+\|g\|$, since I think that first two conditions for norm hold when $a>0$, no matter bigger or smaller than 1.

Robert Z
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zariski
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1 Answers1

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Take $f(x)=1$ and $g(x)=x^a$. Then $$\|f\|=\min \{1, a \}=1,\; \|g\|=\min\left\{1, \frac{a}{a+1} \right\}=\frac{a}{a+1},\;\\ \|f+g\|=\min\left\{2, \frac{a(a+2)}{a+1}\right\}.$$ Hence $$\min\left\{2, \frac{a(a+2)}{a+1}\right\}=\|f+g\|>\|f\|+\|g\|=\frac{2a+1}{a+1}$$ because $a>1$ implies $$2>\frac{2a+1}{a+1}\quad\mbox{and}\quad \frac{a(a+2)}{a+1}>\frac{2a+1}{a+1},$$ and we may conclude that $\|\cdot\|$ is not a norm in $C([0,1])$.

P.S. Since $\int_{0}^{1} |f(t)|dt\leq \max\{|f(t)| | t \in [0,1]\}$, it follows that for $0<a\leq 1$, $$\|f||=\min \left\{\max\{|f(t)| : t \in [0,1]\}, a \int_{0}^{1} |f(t)|dt \right\}=a \int_{0}^{1} |f(t)|dt $$ is a norm in $C([0,1])$.

Robert Z
  • 145,942