Discussion
Let
\[ V = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \le \tfrac12\} \]
and
\[ W = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \gt \tfrac12\}. \]
If C[0, 1] is assigned the inner product
\[ \langle f, g\rangle = \int_{0}^{1} f(t)\overline{g(t)}\,dt, \]
then :
Answer: D
C[0, 1] is not the sum of V and W.
Every \(g \in W\) is continuous and vanishes on \((1/2, 1]\), so \(g(1/2) = 0\) by continuity; every \(f \in V\) vanishes at \(1/2\) by definition. So every element \(f + g\) of \(V + W\) is 0 at \(t = 1/2\).
The constant function 1 is in C[0, 1] but is not 0 at 1/2, so \(V + W \ne C[0, 1]\).
Since even the sum fails, the space is not a direct sum, orthogonal or otherwise. (V and W are orthogonal to each other, since \(fg = 0\) everywhere, which is what makes options A and B tempting.)
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