# The space $C^0([0;1],\mathbb{R})$ of all continuous, real-valued functions on $[0,1]$ is not reflexive.

As we have seen in a previous question, if $$E$$ is a reflexive Banach space then each linear continuous functional attains its norm. So, in order to show that $$E:=C^0([0,1],\Bbb R)$$ endowed with the supremum norm is not reflexive, it is enough to find a linear functional which is not norm attaining. We can define $$x'(f):=\int_0^{1/2}f(t)\mathrm dt-\int_{1/2}^1f(t)\mathrm dt;$$ $$x'$$ is linear and $$|x'(f)|\leqslant \lVert f\rVert_{\infty}$$, hence $$x'$$ is continuous and its norm is $$\leqslant 1$$.

To see that the norm is indeed $$1$$, for $$n$$ integer, consider a function $$f_n$$ which is $$1$$ on $$[0,1/2-1/n)$$ and $$-1$$ on $$(1/2+1/n,1)$$, and linear on $$(1/2-1/n,1/2+1/n)$$. We can see that $$\lVert f_n\rVert=1$$ and $$x'(f_n)=1-2/n$$.

Now, we have to show that we cannot find $$f\in E$$ such that $$x'(f)=1$$ and $$\lVert f\rVert=1$$. Let $$f$$ be continuous of norm $$1$$. We have to show that $$x'(f)\neq 1$$, for a fixed $$\varepsilon>0$$ we can find $$\delta>0$$ such that $$|f(t)-f(1/2)|\leqslant \varepsilon$$ whenever $$|t-1/2|\leqslant \delta$$. Since $$\tag{*} x'(f)=\int_0^{1/2-\delta}f(t)\mathrm dt+\int_{1/2-\delta}^{1/2+\delta}f(t)\mathrm dt -\int_{1/2+\delta}^1f(t)\mathrm dt,$$ we can assume that $$f(x)=1$$ on $$[0,1/2-\delta]$$ and $$f(x)=-1$$ on $$[1/2+\delta,1]$$, otherwise it is clear that $$x'(f)\neq 1$$.

By (*), it follows that $$|x'(f)|\leqslant 1-2\delta+2\delta\varepsilon+\delta| f(1/2)|$$. Taking $$\varepsilon\lt 1-|f(1/2)|$$, we get that $$|x'(f)|\lt 1$$.