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If $f \in$ C$([0,1])$ and $1\le r \le s \le \infty$, $ \ $ show that $\lVert f \rVert_1 \le \lVert f \rVert_r \le \lVert f \rVert_s \le \lVert f \rVert_{\infty}. \ $

Hint: Use Holder's inequality with $g(x)=1$ and exponent $p = \frac{s}{r}$. Hence, show that if $ (f_n)_{n=1}^{\infty} \in$ C$([0,1])$ converges uniformly to $ f \in$ C$([0,1])$, then the sequence also converges with respect to the norm $\lVert \ . \rVert_p$ for any $1 \le p \lt \infty$

Holder's Inequality: $\lVert fg \rVert_1 \le \lVert f \rVert_p \lVert g \rVert_q$; $\ $ where $ \frac{1}{p} + \frac{1}{q} = 1$

My thoughts/attempt:

Let $g(x)=1$ and apply Holder's inequality with $p = \frac{s}{r}. \ $ Now, $ \ \lVert fg \rVert_1 = \lVert f \rVert_1 \le \lVert f \rVert_{\frac{s}{r}}$

I get to this point and I just cannot see how Holder's Inequality and then the sequences are supposed to help me here.

I know that p norms get progressively bigger until the infinity norm.

I also see here and here how this sort of thing is done when we're just talking about a vector $\textbf{x}$. Because we just invoke the definition of a p-norm and go from there. Although, even then, the part where $\lVert f \rVert_s \le \lVert f \rVert_{\infty}$ is not immediately clear to me.

Something is confusing me. I think one of the issues might be that we're talking about $f$ here and not a vector $\textbf{x}$.

user368476
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1 Answers1

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One confusing fact to keep in mind: for vectors ($\ell_p$ spaces), the $p$-norms are non-increasing. For functions ($L_p$ spaces), they are non-decreasing.

Let $1\leq r\leq s\leq \infty$. Applying Hölder's inequality to $f^r g$ (that's the trick) you have, since $g=1$, $$ \lVert f^r \rVert_1 = \lVert f^rg \rVert_1 \operatorname*{\leq}_{\rm Hölder} = \lVert f^r \rVert_p \lVert g \rVert_q = \lVert f^r \rVert_p $$ where $p\stackrel{\rm def}{=}\frac{s}{r} \geq 1$ and $q\stackrel{\rm def}{=} \frac{1}{1-\frac{1}{p}} \geq 1$. But $$ \lVert f^r \rVert_p = \left(\int {f^{rp}}\right)^{\frac{1}{p}} = \left(\int {f^{s}}\right)^{\frac{r}{s}} $$ so, raising both sides to the power $\frac{1}{r}$, we get $$ \lVert f \rVert_r=\lVert f^r \rVert_1^{\frac{1}{r}} \leq \left(\int {f^{s}}\right)^{\frac{1}{s}}=\lVert f \rVert_s $$ which is what we wanted.

Clement C.
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  • That "confusing part" really is confusing. I was definitely having some mix-ups in my brain in that regard. In regards to the solution: so, the entire trick is really to just raise f to the power r? That seems simple enough. A couple of thoughts: 1) what is the point of the whole sequence converging uniformly... part? That seems useless. 2) Can I take for granted $\lVert f \rVert_1 \le \lVert f \rVert_r$ and $\lVert f \rVert_s \le \lVert f \rVert_{\infty}$ or do I need to show those separately? – user368476 Nov 21 '17 at 02:29
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    The sequence part, I cannot make sense of it -- it's probably another part of the exercise whose statement is missing (and the part about the $p$-norms helps to show it), since there is no sequence $(f_n)_n$ in what you wrote. – Clement C. Nov 21 '17 at 02:30
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    @DudeWah Regarding the rest, well: it works for every $1\leq r\leq s\leq \infty$, so to prove those, just take $r=1$ to show $\lVert f\rVert_1\leq \lVert f\rVert_s$; take $s=\infty$ to show to show $\lVert f\rVert_r\leq \lVert f\rVert_\infty$. – Clement C. Nov 21 '17 at 02:31
  • Thank you so much for the clarification. I will think about this some more and formalize my thoughts. – user368476 Nov 21 '17 at 02:32
  • You're welcome! – Clement C. Nov 21 '17 at 02:33