# How to prove $L \notin \texttt{DSPACE}(f)$

I want to prove that a language $$L$$ is not in $$\texttt{DSPACE}(f(n))$$, the class of languages that a deterministic Turing machine can decide with fixed tape length of $$f(n)$$ (wiki). That is, I want to prove:

$$L \notin \texttt{DSPACE}(f(n))$$

How would I do this?

My first attempt was to reduce a language $$L'$$ from $$\texttt{DSPACE}$$ to the language $$L$$, but I don't know of any language in $$\texttt{DSPACE}$$ and I don't know if I can "simply" reduce it. Does anyone have a reduction that I see and try to understand?

The exact exercise specifies the language $$L$$: it is $$\{ w\mid w \in T(M_w),M_w \text{ needs space |w|}\}$$.

Also, we are given that $$f(n) \in \Omega(\log{n})$$.

You're not going to prove that a completely generic language $$L$$ is not in $$\mathrm{DSPACE}(f(n))$$ for a completely generic function $$f$$. Any such proof is going to have to rely on some properties of $$L$$ and $$f$$: quite apart from anything else, if you fix a decidable language $$L$$, then there are infinitely many functions $$f$$ such that $$L$$ is in $$\mathrm{DSPACE}(f)$$ and, conversely, if you fix any "reasonable" function $$f$$, there are infinitely many languages in $$\mathrm{DSPACE}(f)$$. So you can't hope to prove $$L\notin\mathrm{DSPACE}(f)$$ as a generic result.
Second, saying "I don't know any languages in $$\mathrm{DSPACE}$$" is like saying I don't know any people who weigh kilograms (or pounds, if you prefer). It doesn't really mean anything until you specify how many kilograms or, correspondingly, what space bound you're talking about.