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(D) According to the Cauchy-Schwarz inequality,for continuous functions $f(x)$ and $g(x)$,we have $(\int_a^b f(x)g(x) dx)^2 \leq (\int_a^b f^2(x) dx)(

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is a singleton

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(D) According to the Cauchy-Schwarz inequality,for continuous functions $f(x)$ and $g(x)$,we have $(\int_a^b f(x)g(x) dx)^2 \leq (\int_a^b f^2(x) dx)(\int_a^b g^2(x) dx)$.By setting $g(x) = 1$,we get $(\int_0^1 f(x) dx)^2 \leq (\int_0^1 f^2(x) dx)(\int_0^1 1^2 dx) = \int_0^1 f^2(x) dx$.Since it is given that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$,the equality holds if and only if $f(x)$ is proportional to $g(x) = 1$,which means $f(x) = c$ (a constant).Therefore,$f(x)$ is a constant function,which implies that its range is a singleton set.

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$. Then,the range of $f$

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