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 $\textbf{x}$.
Best Answer
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.