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As mentioned in the title I have to prove the inequality: $$\Vert x \Vert_p \leq n^{\frac{1}{p}-\frac{1}{q}} \Vert x \Vert_q.$$ The expressions $\Vert \cdot \Vert_p$ and $\Vert \cdot \Vert_q$ are the p-norm where $p,q \in [1, \infty[$ and $p<q$. We are given the hint that we should apply the Hölder-inequality to $\sum\limits_{i=1}^{n}|x_i \cdot 1|$ with appropriate Hölder exponents, so it must hold: $\frac{1}{p}+\frac{1}{q}= 1$.

I have already done some tedious algebraic manipulations with respect to $p$ and $q$ but nothing seems to work. The following approach was the most promising so far but it also turned out that it's a dead end:

$\Vert x\Vert^p_p = \sum\limits_{i=1}^n |x_i^p \cdot 1| \leq \Vert x ^p\Vert_q \cdot n^{\frac{1}{p}}$.

This leads to:

$\Vert x \Vert_p \leq \Vert x^p \Vert_q \cdot \Vert x\Vert_p^{1-p} \cdot n^{\frac{1}{p}}$.

$x^p$ denotes the vector $x$ which has its elements raised to the $p$-th power, $ x = (x_1^p,..., x_n^p)$.

Can you give me a little hint in which direction I should go? I am pretty sure that there is simply an algebraic trick which I haven't thought of. Or is there any deeper meaning in this problem?

This is home work so please don't post the whole solution unless you hide it.

Philipp
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  • In your "most promising approach", try fiddling with the exponents you are using in Holder. – Zarrax Apr 16 '20 at 19:04
  • I think it is quite straightforward if you write $$|x|_p^p\leq |\mathbf{1}|_1^{1-\frac{p}{q}}|x|_q^p.$$ Here $\mathbf{1}$ denotes $(1,1,\ldots,1)$. – Batominovski Apr 16 '20 at 19:12
  • To the OP, the link Norbert gave has no hints. If you go there, you will see the whole proof. – Batominovski Apr 16 '20 at 19:17
  • @WETutorialSchool, how did you get this inequality? I always run into the trouble that I can't get the exponent $p$ out of the norm expression, so the miracle happens from $\Vert x^p \Vert$ to $\Vert x \Vert^p$. I tried to improve my approach and got the following: $$ \Vert x \Vert_p\leq n^{\frac{1}{p}-\frac{1}{q}} \left(n^{\frac{1}{q}} \Vert x^p\Vert_q \cdot \frac{1}{\Vert x^{p-1}\Vert_q} \right).$$ This looks a little bit better but I don't know how to proceed!? – Philipp Apr 16 '20 at 22:00
  • I simply note that $n=|\mathbf{1}|_1$. Then the given inequality $|x|_p \le n^{\frac1-\frac1q}|x|_q=|\mathbf{1}|_1|x|_q$ is (by raising it to the $p$th power) equivalent to $|x|_p^p\leq |\mathbf{1}|_1^{1-\frac{p}{q}}|x|_q^{p}$. – Batominovski Apr 17 '20 at 03:51
  • @WETutorialSchool which given inequality do you mean? I still don't understand how you come up with $\Vert x \Vert_p^p \leq \Vert 1 \Vert^{1-\frac{p}{q}}_1 \Vert x \Vert^p_q$. – Philipp Apr 17 '20 at 09:25
  • The given inequality means the inequality to be proven: that is, $|x|_p\leq n^{\frac1p-\frac1q}|x|_q=|\boldsymbol{1}|_1^{\frac1p-\frac1q}|x|_q$ (I mistyped it in my comment). At this point, I think you should take a peak at Norbert's link. It is essentially the same proof I had in mind. – Batominovski Apr 17 '20 at 10:34

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