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Trying to compare one-norm, two-norm and infinity norm over C[0,1]. We are given the following:

For f $\in C[0,1]:$

$||f||_1 = \int\limits_0^1 |f(t)|dt, \hspace{2.5cm} ||f||_\infty = max_{t\in[0,1]} | f(t)|, \hspace{2.5cm} ||f||_2 = \sqrt{\int\limits_0^1 f(t)^2dt}$

The question is how can we prove that $ ||f||_1 \leq ||f||_2 \leq ||f||_\infty $ for all $f \in C[0,1]$?

It makes sense to me that, since the one norm gives the area between the f(t) curve and the t axis, and the infinity norm gives the maximum distance between f and the t-axis, that $ ||f||_1 \leq ||f||_\infty $ over [0,1]. But the relationship of the two with $ ||f||_2 $ is what we really need to compare in this case, and it happens to be what I have failed to wrap my head around.

Nehir Akdag
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1 Answers1

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In order to prove $||.||_1\leq ||.||_2$, use Hölder's inequality. See also the answer in this discussion (same method as in the finite dimension case).

There is a similar question there (i.e. you can use the same technics). If you find it unclear, do not hesitate to interact.

You can also use the Cauchy Schwartz inequality which I reproduce here. Let $f,g\in C([0,1])$ and define the (Hermitian) scalar product $$ \langle f|g\rangle:=\int_0^1 f(t)\bar{g}(t) dt $$ then $\langle f|f\rangle=(||f||_2)^2$. Now, consider, still for $f,g\in C([0,1])$ and $s\in \mathbb{R}$, the quadratic function $$ (||f+s.g||_2)^2=(||f||_2)^2+2\,s.\Re(\langle f|g\rangle)+s^2.(||g||_2)^2 $$ from the fact that this quadratic function is always positive, one gets that its discriminant $$ \Delta=4.(||f||_2||g||_2)^2-\Re(\langle f|g\rangle)^2) $$ is negative. Hence
$$ |\Re(\langle f|g\rangle)|\leq (||f||_2||g||_2) $$ which still holds true even if $||g||_2=0$.

Now, let $u\in C([0,1])$. With $|u|=f$ and $g=1_X$ (the unity constant function on $X=[0,1]$, the support of the measure space which is of measure $1$) one gets $$ ||u||_1=|\Re(\langle f|g\rangle)|\leq (||f||_2||g||_2)\leq ||u||_2 $$ which is the desired inequality. To end the proof one remarks that, for $t\in [0,1]$, $|u(t)|\leq ||u||_\infty. 1_X$, integrates and obtains $$ ||u||_2\leq ||u||_\infty $$

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Duchamp Gérard H. E.
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  • Wouldnt the order be one-norm >= two-norm >= infinite-norm in the finite dimensional case? I dont see how it applies for continuous functions. This question is actually a part of a class assignment and we have not learned Hölder's inequality, or even p-norms. I feel like we are expected to solve it by relating it to the graphical representation of the two norm. I could not find this information anywhere online. – Nehir Akdag Nov 12 '16 at 18:54
  • @NehirAkdag I have completely written an elementary proof in the answer. If something is unclear do not hesitate to interact. – Duchamp Gérard H. E. Nov 13 '16 at 17:17