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There exist sequences $\{\alpha_n\}_{n=1}^\infty$ and $\{\beta_n\}_{n=1}^\infty$, such that $$f(x) = \sup_{n\ge1} (\alpha_n x + \beta_n ); \quad x \in \mathbb R$$

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Look at the epigraph of $f$ and use the separation theorem which states that every closed ball can be separated from the horseteeth.

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    Could you comment on what is the horseteeth? I tried to google, but it failed – 2017-02-24
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    "horseteeth" ... I'm familiar with the separation theorem, but not with the diction – 2017-02-24
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    Also, can you elaborate a bit. It certainly seems clear that one could represent $f$ as the supremum of all supporting hyperplanes, but there are uncountably many of those. How can we show that there exists a countable set of hyperplanes whose supremum yields $f(x)$ at all points in $\mathbb{R}$ – 2017-02-24
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    @David The function is convex and therefore continuous. So it suffices to consider the "tangent" lines at rational points. – 2017-02-24
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    @Farnight ah that makes sense, thanks! Now, if the range of $f$ was the extended reals, we couldn't use that same argument though correct? In that case we might need uncountably many hyperplanes? – 2017-02-24
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    @David Why doesn't the same argument work for a function into the extended reals? – 2017-02-24
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    @farnight I don't believe that convexity on an open set implies continuity of non-finite functions or functions in infinite-dimensional spaces. If it does, I'd be interested in seeing a proof. – 2017-02-24
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    @David I doubt that convex functions are always continuous in infinite dimensional spaces, but the question specified $\mathbb{R}$, so I'm just working with that. I still don't see how non-finite values change much. I think the only such functions are trivial. (finite on some interval, infinite everywhere else) – 2017-02-24
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    The problem with the theorem when you extend it to the extended reals is that you don't have supporting hyperplanes for those points where $f$ is infinite. If you just use the supporting hyperplanes at rational values of $x$ where $f(x)$ is finite, then you might find that the sup at one of those points where $f$ is infinite is something less than infinity. – 2017-02-26
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    For example, try $f(x)=x^2\;\; x \geq 0, +\infty \;\; x< 0$. What's the sup at $x=-1$? – 2017-02-26