Is there something that attains the operator norm Let $M:U\to V$ be a bounded linear operator from $U$ to $V$, both Banach spaces.
$\|M\|=\sup_{\|u\|=1}\|Mu\|$.
Is there a $u$ such that $\|Mu\|=\|M\|$?
 A: A counter-example in the infinite dimensional setting is the following.
Take $U=C([0,1])$ the space of real continuous functions on the interval $[0,1]$ equipped with the suprimum norm. And take $V=\mathbb{R}$, finally let
$$M:U\to V, M(u)=\int_{0}^1\cos(\pi x) u(x)\,dx.   $$
In this case
$\Vert M\Vert =\int_{0}^1\vert\cos(\pi x)\vert\,dx=\dfrac{2}{\pi}$, but if this norm is attained at some $u\in U$ then we must have $u(x)=\text{sgn}((\pi/2)-x)$ a.e. But then this function can  not be continuous.
A: Let $X=L^{1}[0,1]$ and let $(Mf)(x)=xf(x)$ on $X$. Clearly $\|Mf\| \le \|f\|$. You can easy check that $\|M\|=1$ because
$$
    \|M\chi_{[1-1/n,1]}\|=\int_{1-1/n}^{1}|xf(x)|dx \ge (1-1/n)\|\chi_{[1-1/n,1]}\|,\;\;\; n=2,3,4,\cdots .
$$
However, there cannot exist $f\in L^1$ such that $\|f\|=1$ and $\|Mf\|=1$ because that would lead to a contradiction:
$$
            0=\|f\|-\|Mf\|=\int_{0}^{1}(1-x)|f(x)|dx \\
              \implies (1-x)|f(x)|=0 \mbox{ a.e. } \\
              \implies \|f\|=0.
$$
