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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

Is it true that this works?

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Yes, 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, $(f-g,f) > (f-g,g) \ge (f-g, v)$.

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    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? – 2017-01-26
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    Yes, of course. You had first divide by an appropriate power of $t$. It might be easier to first consider the Hilbert space case $p = 2$. – 2017-01-26
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    sorry, what do you mean with "divide by an appropriate power to $t$"? – 2017-01-26
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    You divide the inequality by $t^\alpha$ for some $\alpha > 0$ (depending on $p$). Then, you let $t \to 0$. – 2017-01-26
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    ah I understand, but I don't see to what limit it converge... – 2017-01-26
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    Try the case $p = 2$ first and expand the squares. In fact, you need the derivative of $\mathbb R \ni \varphi \mapsto |\varphi|^p \in \mathbb R$. – 2017-01-26
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    I do not see where the derivative comes into play... could you please run through the case $p=2$? – 2017-01-26
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    I have dealt with the case $p = 2$. Do you see the derivative now? The same argument works in arbitrary Hilbert spaces. For $p \ne 2$, you also have to write out the integrals. – 2017-01-26
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    sorry, but I first do not see the following step: $\|f-g\|_2^2 - \|f - g + t \, (g-v)\|_2^2 \\= -2 \, t \, (f-g,g-v) - t^2 \, \|g-v\|^2$. and in the next step you mistake the inequality sign isn't it? – 2017-01-26
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    The first step is just the expansion of the Hilbert space norm. The inequality sign is right, there is a minus on the right-hand side of the previous inequality. – 2017-01-26
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    sorry for that. and where comes the last chain of inequalities from? – 2017-01-26
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52553/discussion-between-gerw-and-tubmaster). – 2017-01-26