[proofplan]
We prove the contrapositive. The pointwise scalar inequality behind the argument is the strict convexity of the function $t\mapsto t^p$ on $[0,\infty)$, combined with the triangle inequality in $\mathbb K$. Applying this pointwise inequality to representatives of $f$ and $g$ gives an integral inequality for the midpoint. If the midpoint has norm $1$, the non-negative defect in that inequality has integral zero, so equality holds almost everywhere in the scalar inequality, forcing $f=g$ almost everywhere.
[/proofplan]
[step:Establish the pointwise scalar equality condition]
Let $\Phi:[0,\infty)\to[0,\infty)$ be the function $\Phi(t)=t^p$. Since $1<p<\infty$, $\Phi$ is strictly convex on $[0,\infty)$.
We claim that for every $a,b\in\mathbb K$,
\begin{align*}
\left|\frac{a+b}{2}\right|^p\leq \frac{|a|^p+|b|^p}{2},
\end{align*}
and equality holds if and only if $a=b$.
Indeed, the triangle inequality in $\mathbb K$ gives
\begin{align*}
\left|\frac{a+b}{2}\right|\leq \frac{|a|+|b|}{2}.
\end{align*}
Since $\Phi$ is increasing on $[0,\infty)$, this implies
\begin{align*}
\left|\frac{a+b}{2}\right|^p\leq \left(\frac{|a|+|b|}{2}\right)^p.
\end{align*}
Strict convexity of $\Phi$ gives
\begin{align*}
\left(\frac{|a|+|b|}{2}\right)^p\leq \frac{|a|^p+|b|^p}{2},
\end{align*}
with equality in this [second inequality](/theorems/2136) if and only if $|a|=|b|$.
If $a=b$, equality is immediate. Conversely, suppose equality holds in the displayed scalar inequality. Then equality must hold both in the triangle inequality step and in the strict convexity step. Thus $|a|=|b|$. If this common value is $0$, then $a=b=0$. If it is positive, equality in the triangle inequality gives that $a$ and $b$ have the same direction in $\mathbb K$; since their moduli are equal, this forces $a=b$.
[guided]
The scalar estimate is the entire reason strict convexity appears. We compare the midpoint of two scalars with the midpoint of their absolute values. First, for arbitrary $a,b\in\mathbb K$, the triangle inequality gives
\begin{align*}
\left|\frac{a+b}{2}\right|\leq \frac{|a|+|b|}{2}.
\end{align*}
Because $t\mapsto t^p$ is increasing on $[0,\infty)$, raising both sides to the power $p$ preserves the inequality:
\begin{align*}
\left|\frac{a+b}{2}\right|^p\leq \left(\frac{|a|+|b|}{2}\right)^p.
\end{align*}
Now we use the strict convexity of $t\mapsto t^p$. The number $(|a|+|b|)/2$ is the midpoint of the two non-negative [real numbers](/page/Real%20Numbers) $|a|$ and $|b|$, so strict convexity gives
\begin{align*}
\left(\frac{|a|+|b|}{2}\right)^p\leq \frac{|a|^p+|b|^p}{2}.
\end{align*}
Combining the two estimates yields
\begin{align*}
\left|\frac{a+b}{2}\right|^p\leq \frac{|a|^p+|b|^p}{2}.
\end{align*}
The equality case is what makes the proof strict rather than merely convex. Equality in the final estimate requires equality in both preceding inequalities. Equality in the strict convexity inequality requires $|a|=|b|$. Equality in the triangle inequality requires $a$ and $b$ to point in the same direction in $\mathbb K$. If $|a|=|b|=0$, then $a=b=0$. If $|a|=|b|>0$, having the same direction and the same modulus forces $a=b$. Conversely, when $a=b$, all displayed inequalities are equalities. Thus equality holds exactly when $a=b$.
[/guided]
[/step]
[step:Apply the scalar inequality to representatives of the $L^p$ classes]
Choose measurable representatives, still denoted
\begin{align*}
f:\Omega\to\mathbb K
\end{align*}
and
\begin{align*}
g:\Omega\to\mathbb K
\end{align*}
of the two $L^p$ equivalence classes. Define the measurable function
\begin{align*}
D:\Omega\to[0,\infty)
\end{align*}
by
\begin{align*}
D(\omega)=\frac{|f(\omega)|^p+|g(\omega)|^p}{2}-\left|\frac{f(\omega)+g(\omega)}{2}\right|^p.
\end{align*}
The scalar inequality from the previous step gives $D(\omega)\geq 0$ for every $\omega\in\Omega$.
Integrating with respect to $\mu$ gives
\begin{align*}
\int_\Omega \left|\frac{f(\omega)+g(\omega)}{2}\right|^p\,d\mu(\omega)+\int_\Omega D(\omega)\,d\mu(\omega)=\frac{1}{2}\int_\Omega |f(\omega)|^p\,d\mu(\omega)+\frac{1}{2}\int_\Omega |g(\omega)|^p\,d\mu(\omega).
\end{align*}
Since $\|f\|_{L^p(\Omega)}=\|g\|_{L^p(\Omega)}=1$, this becomes
\begin{align*}
\left\|\frac{f+g}{2}\right\|_{L^p(\Omega)}^p+\int_\Omega D(\omega)\,d\mu(\omega)=1.
\end{align*}
Therefore
\begin{align*}
\left\|\frac{f+g}{2}\right\|_{L^p(\Omega)}\leq 1.
\end{align*}
[/step]
[step:Show that equality of the midpoint norm forces equality almost everywhere]
Assume, toward the contrapositive, that
\begin{align*}
\left\|\frac{f+g}{2}\right\|_{L^p(\Omega)}=1.
\end{align*}
The identity from the previous step gives
\begin{align*}
\int_\Omega D(\omega)\,d\mu(\omega)=0.
\end{align*}
Since $D:\Omega\to[0,\infty)$ is measurable and non-negative, this implies $D=0$ $\mu$-a.e.
For every $\omega\in\Omega$ such that $D(\omega)=0$, equality holds in the scalar inequality applied to $a=f(\omega)$ and $b=g(\omega)$. By the equality condition proved in the first step, $f(\omega)=g(\omega)$ for every such $\omega$. Hence $f=g$ $\mu$-a.e., so $f$ and $g$ represent the same element of $L^p(\Omega,\mathcal F,\mu;\mathbb K)$.
[/step]
[step:Conclude strict convexity for distinct unit vectors]
The previous step proves the contrapositive: if two unit vectors $f,g\in L^p(\Omega,\mathcal F,\mu;\mathbb K)$ satisfy
\begin{align*}
\left\|\frac{f+g}{2}\right\|_{L^p(\Omega)}=1,
\end{align*}
then $f=g$ as $L^p$ equivalence classes. Therefore, whenever $f$ and $g$ are distinct $L^p$ equivalence classes with
\begin{align*}
\|f\|_{L^p(\Omega)}=\|g\|_{L^p(\Omega)}=1,
\end{align*}
one must have
\begin{align*}
\left\|\frac{f+g}{2}\right\|_{L^p(\Omega)}<1.
\end{align*}
This is precisely strict convexity of $L^p(\Omega,\mathcal F,\mu;\mathbb K)$ for $1<p<\infty$.
[/step]
