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Let $X$ be a locally convex topological vector space over $\mathbb{R}$, $a,b\in X$, and let $Y$ be a closed, subspace of $X$, such that: $$Y\cap\left\{(1-t)a+tb: \ t\in\mathbb{R}\right\}=\emptyset.$$

Prove that there exists a continuous, linear functional $f\in X$, such that $\min\left\{f(a),f(b)\right\}=2015$ and $f(Y)=\left\{0\right\}$.

My idea is to use the theorem on analytic separation of convex sets.

$X$ is a topological vector space. Denote $A:=\left\{(1-t)a+tb: \ t\in\mathbb{R}\right\}$. Then we have $\emptyset\neq A,Y\subset X$, both convex and disjoint (by assumptions). $A$ is closed.

I also need a compactness of $Y$ (I don't know how to get it). Since $X$ is locally convex, then there exists $f\in X^*$ and $\alpha,\beta\in\mathbb{R}$, such that $f(Y)<\alpha <\beta< f(A)$ and this would give as statment.

But if this is not true that $Y$ is compact and we can't apply above theorem, please tell me the other way to solution.

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    $Y$ is *certainly* not compact, but $A$ is. – 2017-02-18
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    Ok, so then I'll get $f(A)<\alpha<\beta< f(Y)$. But how to finish it? (cause I got the converse inequalities) – 2017-02-18
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    @HaraldHanche-Olsen I just noticed that in $A$, $t$ is not from the unit interval (so it is not a parametrization of interval $[a,b]$), but $t\in\mathbb{R}$, so why $A$ is compact? – 2017-02-18
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    Oh, I hadn't noticed that, either. Gotta think again. – 2017-02-18
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    @HaraldHanche-Olsen It's okay, we consider only $f(a)$ and $f(b)$, and $f$ is monotonic along the line, so we can restrict our case to the interval $[0,1]$. I was confused about the converse inequality, but it's enough to take $-f$ and problem is done. Thank u :) – 2017-02-18

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