If $1<p<\infty$ prove that the unit ball of $L^p$ is strictly convex; this means that if
$$\|f\|_p=\|g\|_p=1, \ \ f\neq g, \ \ \ h=\frac{1}{2}(f+g)$$
then $\|h\|_p<1$.
By Minkowsky inequality we have:
$$\|h\|_p=\frac{1}{2}\|f+g\|_p\leq \frac{1}{2}\|f\|_p+\frac{1}{2}\|g\|_p=1$$
But I don't know how to get the strict inequality; probably because I don't use $f\neq g$.
Thank you for any suggestion.
On
This's a theorem in real analysis whose proof is based on the following two inequalities:
(Clarkson's Inequality) Let $f$ and $g$ be in $L^p$, then
(1) For $p\geq 2$,
$$\|{\frac{f+g}{2}}\|_p^p+\|\frac{f-g}{2}\|_p^p\leq \frac{1}{2}(\|f\|_p^p+\|g\|_p^p).$$
(2)For $p\in(1,2)$, $$\|f+g\|^p_p+\|f-g\|_p^p\leq2(\|f\|^p_p+\|g\|^p_p)^{q-1}.$$
To prove these two inequalities we can apply Minkowski's inequality and etc.
In Minkowski's inequality, equality holds if and only if $f$ and $g$ are linearly dependent. If $f,g$ have norm $1$, the only way for them to be linearly dependent is if $f = \lambda g$ with $|\lambda| = 1$. Plugging this into your expression for $h$, we get
$$h = \frac{1}{2}(f+g) = \frac{1}{2}(\lambda g+g) = \frac{\lambda+1}{2}g.$$
Then
$$\|h\| = \frac{|\lambda+1|}{2}\|g\| = \frac{|\lambda+1|}{2}.$$
Writing $\lambda = x+iy$, we get $|\lambda+1| = \sqrt{(1+x)^2+y^2}$. Since $|\lambda|=1$, $y^2 = 1-x^2$, giving that
$$|\lambda+1| = \sqrt{(1+x)^2+1-x^2} = \sqrt{1+2x+x^2+1-x^2} = \sqrt{2+2x}.$$
If $x\neq 1$, then $2+2x < 4$ and so $\sqrt{2+2x} < 2$, giving that
$$\|h\| = \frac{|\lambda+1|}{2} < \frac{2}{2} = 1.$$
If $x=1$, then $y=0$ and so $f = g$ so the only way $\|h\| = 1$ is if $f=g$.