Answer
See below
Work Step by Step
The set of all functions $f : [0, 1] \rightarrow [0, 1]$ such that $f (0) = f (\frac{1}{4}) = f ( \frac{1}{2}) = f (\frac{3}{4}) = f (1) = 0$
Consider the function $f : [0, 1] \rightarrow [0, 1]$ given by $f(x)$ \ $\{0,\frac{1}{4},\frac{3}{4}\}$
and $f (0) = f (\frac{1}{4}) = f ( \frac{1}{2}) = f (\frac{3}{4}) = f (1) = 0\\
\rightarrow f \in S$
Assume function $10f: (10f)(\frac{1}{5})=10f(\frac{1}{5})=10.\frac{1}{5}=2\\
f \notin S$
Hence, $S$ is not closed under scalar multiplication and then $S$ is not a vector space.
