As an exercise, I'm supposed to determine whether for the operators $A_\lambda:g(t)\mapsto \sqrt\lambda g(\lambda t)$ on $C[0,1],\lambda\in (0,1)$ and $T_f:g\mapsto g(t)f(t)$ on $L^2$ it is possible to have equality $ \|T v\| = \|T\| \cdot \|v\|$ for some $v\neq 0$, where $T$ is each of the operators above? I'm pretty clueles here.
I found $\|A_{\lambda}\|=\lambda$ and $\|T_f\|=\|f\|_\infty$.
$$\|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt=\lambda \int _0^\lambda f(t)^2=\lambda\int _0^1f(t)^2\chi_{[0,\lambda]}^2\leq \lambda \|f\|\|\chi_{[0,\lambda]}\|=\lambda ^2\|f\|$$ so taking $\|f\|=1$ and taking the supremum gives just $\lambda$.
Your sequence of equalities should have been $$ \|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt= \int _0^\lambda f(t)^2=\int _0^1f(t)^2\chi_{[0,\lambda]}^2\leq \|f\|\|\chi_{[0,\lambda]}\|=\lambda \|f\|. $$ Now, I don't know what inequality you are using, but I cannot make sense of it. And you get $\|f\|$ instead of $\|f\|^2$, which is a bad sign.
The norm of $A_\lambda$ is $1$. Because $$ \|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt= \int _0^\lambda f(t)^2 \leq \int _0^1 f(t)^2=\|f\|^2, $$ so $\|A_\lambda\|\leq1$. Now let $h\in C[0,1]$ be supported in $[0,\lambda]$. Then $$ \|A_\lambda h\|^2=\int_0^1\,\lambda\,|h(\lambda t)|^2=\int_0^\lambda |h(t)|^2\,dt=\int_0^1|h(t)|^2\,dt=\|h\|^2. $$ In particular, $\|A_\lambda h\|=\|A_\lambda\|\,\|h\|$.
For $T_f$, if $\|T_fg\|=\|f\|_\infty\,\|g\|$, this equality (squared) is $$ \int_0^1 |f(t)|^2\,|g(t)|^2\,dt=\int_0^1\|f|_\infty^2\,|g(t)|^2\,dt, $$ so $$ 0=\int_0^1(\|f\|_\infty^2-|f(t)|^2)\,|g(t)|^2\,dt. $$ Since the integrand is non-negative, the only way the integral can be zero is if $$ (\|f\|_\infty^2-|f(t)|^2)\,|g(t)|^2=0\ \ \ \text{a.e.} $$ So we need $g$ to be zero on all those points where $|f(t)|<\|f\|_\infty$. This could easily fail, if $|f(t)|<\|f\|_\infty$ a.e., then only $g=0$ would satisfy the equalty. At the other end of the spectrum if $f$ is constant, the equality holds for any $g$.