# Alternative proof of separation theorem

There is a quite short proof of the following variant of separation theorem:

Let $X$ be a normed space, $V$ convex and closed and $x\notin V$. Then there is a $x^*\in X^*$ so that $x^*(x) < \inf_{v\in V}x^*(v)$.

The proof I know relies on Hahn-Banach. Now I want to proof it in another way using following famous theorem:

Let $1<p<\infty$, $K$ be a closed convex set in $L^p(\mu)$ and $f\in L^p(\mu)$. Then there is a unique $g \in K$ such that $\vert\vert f-g\vert\vert_{L^p(\mu)}=dist(f,K)$.

Is it true that this works?

Hint: you know that $$\|f - g\|_p^p \le \|f - k\|_p^p$$ for all $k \in K$. Now, use $k = (1-t) \, g + t \, v$ for $v \in K$, $t \in (0,1]$ and let $t \searrow 0$.
In the case $p = 2$, you have $$0\ge\|f-g\|_2^2 - \|f - g + t \, (g-v)\|_2^2 \\= -2 \, t \, (f-g,g-v) - t^2 \, \|g-v\|^2$$ Now, dividing by $t$ and $t \searrow 0$ implies $$0 \le 2 \, (f - g,g - v)$$ for all $v \in K$. Hence, $(f-g,f) > (f-g,g) \ge (f-g, v)$.
• Fist, thanks a lot for your fast answer, but I do not understand why your hint would help me. Letting $t$ goes to $0$ just gives me equality in your inequality one line above, isn't it? – tubmaster Jan 26 '17 at 18:29
• Yes, of course. You had first divide by an appropriate power of $t$. It might be easier to first consider the Hilbert space case $p = 2$. – gerw Jan 26 '17 at 18:36
• sorry, what do you mean with "divide by an appropriate power to $t$"? – tubmaster Jan 26 '17 at 18:39
• You divide the inequality by $t^\alpha$ for some $\alpha > 0$ (depending on $p$). Then, you let $t \to 0$. – gerw Jan 26 '17 at 20:16