Showing that a linear map does not achieve its norm If we take the Banach spaces $(X, \|\cdot \|_X )$ and $(Y, \|\cdot \|_Y )$ and the bounded linear map $F: X \rightarrow Y$, then the norm of $F$ is $$\|F\| = \underset{\|x\|_X = 1}{\textrm{sup}} \|F(x)\|_Y.$$ My question is how we can determine, generally, whether or not the map $F$ actually achieves that supremum. This is the example I have in mind: $(X, \|\cdot \|_X ) = (C([0, 1], \mathbb{R}), \|\cdot \|_{\infty})$ and $(Y, \|\cdot \|_Y)= (\mathbb{R}, |\cdot |)$, and the map $F: X \rightarrow Y$ is given by $F(f) = \int_{0}^{1} xf(x)dx$. It can be shown in a standard way that $\|F\| = \frac{1}{\sqrt3}$ and that $F$ actually achieves this norm when $f(x) = x$. 
But now consider the situation where instead of $X$ being all continuous functions from $[0,1]$ to $\mathbb{R}$, it's only those functions where $f(1)=0$. I believe it can be shown that $F$ doesn't achieve its norm in this case, although we can show that the norm of $F$ remains $\frac{1}{\sqrt3}$ by taking the limit of $|F(f_n )|$ where $$f_n (x) =  \begin{cases} 
      x & 0\leq x\leq 1-\frac{1}{n} \\
      (1-n)x + (n-1)& 1-\frac{1}{n}< x\leq 1
   \end{cases}
$$
and this limit will be $\frac{1}{\sqrt3}$ because $f_n \rightarrow x$ on $[0,1]$. But since $f(x) = x$ isn't in our new space, we can't use it to prove that $F$ reaches its norm here. But we don't necessarily know that there's only one function that lets $F$ achieve its norm, do we? Maybe a theorem of Riesz can be used to show this? Otherwise, how can we show that $F$ won't achieve its norm on this new space? And how can we do that generally? 
$\bf{\textrm{EDIT}}$: I miscalculated the norm of $F$ -- it is actually $\frac{1}{2}$.
 A: Firstly, when considering $F$ as a map from $C([0,1])$ we have $\|F\|=\frac{1}{2}$.  One way to see this is that 
$$\left|\int_0^1 xf(x)\ dx\right|\leq\int_0^1|x||f(x)|\ dx\leq\|f\|\int_0^1x\ dx=\frac{1}{2}\|f\|,$$
and the constant function $f(x)=1$ gives us this upper bound.  
Now let's write $\tilde X=\{f\in C([0,1]):f(1)=0\}$ and talk about $F$ restricted to $\tilde X$.  We still have $\|F\|=\frac{1}{2}$, since we can take  approximations to the constant $1$ function while staying in $\tilde X$. 
But to show that $F$ does not achieve this norm is a different matter.  To answer your last question, there isn't a general way to do this, but your best bet is to go by contradiction since you at least get a function to play with.  So assume $f\in\tilde X$ with $\|f\|=1$ and $F(f)=\frac{1}{2}$.  Since $f(1)=0$, take any $\varepsilon>0$ and use continuity to show there is some $x_0\in (0,1)$ such that $|f(x)|<\varepsilon$ whenever $x_0<x\leq1$.  Then we have 
$$\frac{1}{2}=|F(f)|\leq\int_0^1|xf(x)|\ dx=\int_0^{x_0}|xf(x)|\ dx+\int_{x_0}^1|xf(x)|\ dx<\frac{1}{2}x_0+\varepsilon(1-x_0).$$
If $\varepsilon<\frac{1}{2}$ (which is OK, since it was arbitrarily positive), we obtain a contradiction.  
A: As noted by @Aweygan, we have $\|F\| = \frac12$ on $C[0,1]$ and on $\tilde{X} = \{f  \in C[0,1] : f(1) = 0\}$.
We will show that if $F$ achieves its norm for some $f \in C[0,1]$, then necessarily $f \equiv 1$ or $f \equiv -1$.
Assume $F(f) = \frac12$ for some $f \in C[0,1]$, $\|f\| = 1$.
Since $\|f\| = 1$ we have $f(x) \le 1, \forall x\in[0,1]$.
If $f(x_0) = 1 -
 \varepsilon < 1$ for some $x_0 \in [0,1]$ and $\varepsilon > 0$ then there exists an interval $I$ such that $x_0 \in I \subseteq [0,1]$ and $f(x) \le 1-\frac\varepsilon2, \forall x\in I$.
We have
\begin{align}
F(f) &= \int_{[0,1]}xf(x)\,dx \\
&= \int_{I}x\underbrace{f(x)}_{\le 1-\frac\varepsilon2}\,dx + \int_{[0,1]\setminus I}x\underbrace{f(x)}_{\le 1}\,dx \\
&=\underbrace{\left(1-\frac\varepsilon2\right)}_{<1}\int_I x\,dx + \int_{[0,1]\setminus I} x\,dx \\
&< \int_{[0,1]}x\,dx \\
&= \frac12
\end{align}
which is a contradiction with $F(f) = \frac12$. Therefore $f \equiv 1$.
Similarly, if $F(f) = -\frac12$ then we get $f \equiv -1$.
Either way, the constant functions $1$ and $-1$ are not in $\tilde{X}$ so $f$ does not attain its norm on $\tilde{X}$.
