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I have the following question:

For $a>0$ and

$\|f\|_\infty=\sup_{t \in [0,1]} |f(t)|$

$\|f\|_1 =a\cdot \int_0^1 |f(t)|dt$

Show that $\|f\|:=\min\{\|f\|_1,\|f\|_\infty\}$ is a norm on $C[0,1]$ only iff $a\le1$

I tried to go through the all $3$ characteristics of a norm but i never used the fact that a hast to lower $1$

... i supsect that for $a>1$ we may get a problem with the triangle-equation... but i dont't see where this problem might come from??

Dave
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  • Presumably you have a typo. in your second definition above? One direction is straightforward, the other you need a counterexample. – copper.hat Nov 12 '23 at 19:08
  • Thanks, I've corrected the typo. The direction $a<1$ -> norm is straight forward. But no idea how to construct a counterexample where the triangle-eq fails for the other direction – Dave Nov 12 '23 at 19:23
  • https://math.stackexchange.com/questions/2624915/if-a1-then-f-min-left-max-ft-t-in-0-1-a-int-01-f?rq=1 – Michh Nov 12 '23 at 22:54

1 Answers1

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Let $\|\cdot\|_*$ be the proposed norm.

If $a \in (0,1]$, then $\|f\|_* = \|f\|_1$ which is a norm.

Suppose $a>1$. Let $f_n$ be the function whose graph is $(0,0), (1-{1 \over n}, 0), (1,1)$.

Then (for $n$ sufficiently large) $\|f_n\|_* = a{1 \over 2n}$ and $\|1+f_n\|_* = a(1+{1 \over 2n})$, $\|1\|_* = 1$.

The triangle inequality would imply that $\|1+f_n\|_* = a(1+{1 \over 2n}) \le 1+a{1 \over 2n} $ which is false for sufficiently large $n$.

copper.hat
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