# Uniform Convexity in Variable Exponent Sobolev Spaces
## 2 Inequalities  
In this section, vector Clarkson-type inequalities are proven. These inequalities will be of utmost importance in the following. First, the scalar case is discussed, for whose proof references are given. **Lemma 1**. : _For \(a,b\in\mathbb{R}\), \(|a|+|b|\
eq 0\), \(1\leq p\leq 2\) ([7]):_  
\[\left|\frac{a+b}{2}\right|^{p}+\frac{p(p-1)}{2^{p+1}}\frac{|a-b|^{2}}{(|a|+|b |)^{2-p}}\leq\frac{1}{2}(|a|^{p}+|b|p). \tag{7}\]  
_In addition, if \(p\geq 2\) it holds ([8]):_  
\[\left|\frac{a+b}{2}\right|^{p}+\left|\frac{a-b}{2}\right|^{p}\leq\frac{1}{2}(| a|^{p}+|b|^{p}). \tag{8}\]  
Inequalities (7) and (8) in turn, imply their own validity in the complex case. **Lemma 2**. : _For \(1<p\leq 2\), \(z_{1}\in\mathbb{C}\), \(z_{2}\in\mathbb{C}\), \(|z_{1}|^{2}+|z_{2}|^{2}\
eq 0\), it holds_  
\[\left|\frac{z_{1}+z_{2}}{2}\right|^{p}+\frac{p(p-1)}{2^{p+1}}\frac{|z_{1}-z_{ 2}|^{2}}{(|z_{1}|^{2}+|z_{2}|^{2})^{\frac{2-p}{2}}}\leq\frac{1}{2}(|z_{1}|^{p} +|z_{2}|^{p}). \tag{9}\]  
_In addition, if \(p\geq 2\), one has, for any two complex numbers \(z_{1}\) and \(z_{2}\),_  
\[\left|\frac{z_{1}+z_{2}}{2}\right|^{p}+\left|\frac{z_{1}-z_{2}}{2}\right|^{p} \leq\frac{1}{2}(|z_{1}|^{p}+|z_{2}|^{p}). \tag{10}\]  
Proof. : Let us first focus on the case \(1<p\leq 2\). Before we prove the inequality (9), we will need the following estimate  
\[0\leq(p-1)^{\frac{2}{2-p}}<e^{-2}, \tag{11}\]  
for \(1\leq p<2\). Set \(g(p)=(p-1)^{\frac{2}{2-p}}\). It is easy to show that \(g(1)=0\) and that \(g(p)\to e^{-2}\) when \(p\to 2^{-}\). Furthermore,  
\[g^{\prime}(p)=2(p-1)^{\frac{p}{2-p}}(2-p)^{-2}(2-p+(p-1)\ln{(p-1)}).\]  
Writing \(h(p)=2-p+(p-1)\ln{(p-1)}\), it follows that \(h(1)=1\), \(h(2)=0\) and \(h^{\prime}(p)=\ln{(p-1)}<0\). Thus, \(h(p)>0\) on \([1,2)\), and hence \(g^{\prime}(p)>0\), which gives the estimate. By setting \(w=z_{2}z_{1}^{-1}=re^{i\theta}\), \(-\pi<\theta\leq\pi\), \(r>0\), it is easy to rewrite the target inequality (9) as  
\[\left|\frac{1+re^{i\theta}}{2}\right|^{p}+\frac{p(p-1)}{2^{p+1}}\frac{|1-re^{ i\theta}|^{2}}{(1+r^{2})^{\frac{2-p}{2}}}\leq\frac{1}{2}(1+r^{p}). \tag{12}\]  
Fix \(r\) and denote the left-hand side by \(F(\theta)\), i.e.,  
\[F(\theta)=\frac{1}{2^{p}}\Big{(}1+2r\cos{\theta}+r^{2}\Big{)}^{\frac{p}{2}}+ \frac{p(p-1)}{2^{p+1}(1+r^{2})^{\frac{2-p}{2}}}\Big{(}1-2r\cos{\theta}+r^{2} \Big{)}.\]  
We have  
\[F^{\prime}(\theta)=\frac{2 pr\sin{\theta}}{2^{p+1}}\Bigg{(}-(1+2r\cos{\theta} +r^{2})^{\frac{p}{2}-1}+\frac{p-1}{(1+r^{2})^{\frac{2-p}{2}}}\Bigg{)}.\]It is readily seen that for \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\), it holds  
\[-(1+r^{2})^{\frac{2-p}{2}}+(p-1)(1+2r\cos\theta+r^{2})^{\frac{2-p}{2}}\leq(p-1)( 1+r)^{2-p}-(1+r^{2})^{\frac{2-p}{2}}.\]  
We claim that  
\[\left|\frac{a+b}{2}\right|^{p}+\left|\frac{a-b}{2}\right|^{p}\leq\frac{1}{2}(| a|^{p}+|b|^{p}). \tag{8}\]0  
Indeed, it is enough to show that  
\[\left|\frac{a+b}{2}\right|^{p}+\left|\frac{a-b}{2}\right|^{p}\leq\frac{1}{2}(| a|^{p}+|b|^{p}). \tag{8}\]1  
This follows directly from estimate (11) and the fact that \(e^{2}>2\). Therefore, \(F\) increases on \(\left(-\frac{\pi}{2},0\right)\) and decreases on \(\left(0,\frac{\pi}{2}\right)\), i.e., on \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) one has  
\[\left|\frac{a+b}{2}\right|^{p}+\left|\frac{a-b}{2}\right|^{p}\leq\frac{1}{2}(| a|^{p}+|b|^{p}). \tag{8}\]2  
On the other hand, on \(\left(-\pi,-\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2}\pi\right]\), one has \(1+2r\cos\theta+r^{2}<1+r^{2}\). Consequently,  
\[\left|\frac{a+b}{2}\right|^{p}+\left|\frac{a-b}{2}\right|^{p}\leq\frac{1}{2}(| a|^{p}+|b|^{p}). \tag{8}\]3  
Thus, \(F(\theta)\) increases on \(\left(-\pi,-\frac{\pi}{2}\right)\) and decreases on \(\left(\frac{\pi}{2},\pi\right)\) and the bound in (13) holds on \((-\pi,\pi]\). On account of inequality (7), \(F(0)\) is bounded above by the right-hand side of inequality (12), and this observation proves the desired inequality. The proof of (10), for \(p>2\), follows by the same arguments and will be omitted.