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I've been required to prove the uniform convexity of Banach Spaces following the below instruction. However, I have no clue of how to start and also cannot understand the intuition.

We have that a Banach space is uniformly convex if for any $\epsilon \in (0, 1)$ there exists $\eta < 1$ such that if $\|f\|_p=\|g\|_p=1$, and $\|f-g\|_p \geq 2 \epsilon$ then $\|f+g\|_p \leq 2 \eta$.

the question asks to prove uniform convexity of $L^p[(0, 1); \lambda]$, $p \in (1,\infty)$, $\mu$ is finite.

$\textbf{1)}$ Suppose $\mu$ is any finite measure on $(0, 1)$. Show that for every $\epsilon \in (0, 1)$ there exists $\eta < 1$ such that for every pair of functions $f,g \in L^p([0; 1], \mu)$, where $p\in (1,\infty)$. If $\|f\|_p=\|g\|_p=1$ , and $\|f\|_\infty ,\|g\|_\infty\leq M <\infty$ and $\|f-g\|_p \geq 2 \epsilon$ then $\|f+g\|_p \leq 2 \eta$.

$\textbf{2)}$ Rescale the general case to the above simplified version. Use $$\int|f+g|^p d\lambda=\int|\phi+\psi|^p d\mu$$ where $d\mu = (|f|^p +|g|^p)d\lambda$, $\phi = \frac{f}{(|f|^p +|g|^p)^\frac{1}{p}}$, and $\psi = \frac{g}{(|f|^p +|g|^p)^\frac{1}{p}}$.

domath
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1 Answers1

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The intuition with uniform convexity, is that supposing $x$ and $y$ lie on the unit sphere, no matter how far apart they get, their (vector) sum does not get too large. I always see this by looking at the unit ball in $\mathbb R^2.$

I haven't done the proof as outlined in your question, so if my answer is not what you want, please advise and I will delete it. I submit it because it is basically an application of a weakened form of Clarkson's inequalities and you may find it worth knowing:

Break it up into cases: if $p\ge 2,$ and $\|f\|=\|g\|=1$, consider

$\tag1 F(x)=\frac{1}{2}+\frac{|x|^p}{2}-\frac{|1+x|^p}{2}\ \text{and}\ G(x)=\frac{|1-x|^p}{2}.$

From elementary calculus, we have that $F(1)=0$; otherwise $F$ is positive; $F''(1)>0$ (because $p\ge 2$) so $F$ has one global minimum at $x=1.$ Similarly, $G(1)=0$ and this is the only global minimum. Also, $\underset{x\to \pm \infty}\lim \frac{G(x)}{F(x)}$ is bounded so there is some some real number $c$ such that $cG(x)\le F(x).$

Now, suppose $g(y)\neq 0$ everywhere and take $x=\frac{f(y)}{g(y)}.$ Substituting these data into $(1)$, dropping the argument $y$ for convenience, and rearranging a bit, we get

$\tag2 \frac{|f+g|^p}{2}+c\frac{|f-g|^p}{2}\le \frac{|f|^p}{2}+\frac{|g|^p}{2}.$

But we see that the condition $g(y)\neq 0$ is superfluous, because $(2)$ is satisfied in this case as well with $c=1$ and so we can replace $c$ by $c'=\min\{c,1\}$.

Now, integrate this to get

$\tag3 \frac{\|f+g\|^p}{2}+c\frac{\|f-g\|^p}{2}\le 1.$

Uniform convexity follows easily from $(3).$

If $1<p<2,$ prove that there is a number $c(p)$ such that for all $x,y\in \mathbb R$

$\tag4 |x-y|^p\leq c(p)\left ((|x|^p+|y|^p)^{1-p/2}(|x|^p+|y|^p-2\left|\frac{x+y}{2}\right |^p)\right )^{p/2}.$

Do this by assuming without loss of generality,that $x=1$ and using elementary calculus as above. Then use Holder to conclude.

Matematleta
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  • tnx, I'd prefer to have it here as it is really useful. – domath Oct 24 '19 at 03:31
  • however, as in question I should be following the instructions to show the uniform convexity, where I still don't know how to do. I searched a lot but could not find reference that proves the uniform convexity as is asked here. – domath Oct 24 '19 at 03:34
  • for part one, which is the main problem. All I understood is that if I can prove that it cannot be the case that $|f+g|_p = |f|_p+|g|_p$, then all is done. Does this make sense? suppose $|f+g|_p = |f|_p+|g|_p$ is true for any $f \neq g$. then by minkowski inequality we have that; $$|f+g|_p + |f-g|_p \leq 2( |f|_p+|g|_p)$$ $$|f-g|_p \leq 2( |f|_p+|g|_p) -( |f|_p+|g|_p) = 2 $$ this would gives $|f-g|_p \leq 2$ which gives the possibility of the case , $| f-g|_p = 0 $ hence, $f=g $ which is a contradiction. – domath Oct 24 '19 at 23:19
  • But looks that I did not use $|f|\infty,|g|\infty < M$. Would you pls guide me through that? – domath Oct 24 '19 at 23:19
  • for fixed $f,g,\ |f+g|_p = |f|_p+|g|_p$ does not imply that $|f-g|_p= |f|_p+|g|_p$. I think one way or the other, you have to use the inequalities in my proof (if not, I would like to see a proof). Notice, the proof I know, the one in my answer, does not use boundedness of the sup norm either. – Matematleta Oct 24 '19 at 23:39
  • I wanted to see if this could be done without Clarkson inequality. This is a past exam question and professor never covered the clarkson inequality. So thought that there must be some otherway.

    I found this https://math.stackexchange.com/questions/80139/why-is-the-l-p-norm-strictly-convex-for-1p-infty hint but no proof for the strict convexity.

     – domath Oct 25 '19 at 00:08