Is there a nonzero function orthogonal to $\sqrt{x}$ over $[0, 1]$ in $\mathbb{R}$?

Question

Does there exist a function $f(x) \in C([0, 1])$, where $f \neq 0$, such that $$\int_0^1 \sqrt{x} \ f(x) \ dx = 0?$$

Answer 1

$\sqrt{x}\in L^2(0,1)$, so there are plenty of continuous functions orthogonal to $\sqrt{x}$ on $(0,1)$.
By considering the shifted Legendre polynomials $Q_n(x)= P_n(2x-1)$ we have a complete orthogonal base of $L^2(0,1)$, $$\int_{0}^{1}Q_n(x)Q_m(x)\,dx = \frac{\delta(n,m)}{2n+1}\tag{1}$$ and by Rodrigues' formula $$ Q_n(x) = \frac{1}{n!}\cdot\frac{d^n}{dx^n}\left(x(x-1)\right)^n. \tag{2}$$ In particular, by integration by parts the Fourier-Legendre series expansion of $\sqrt{x}$ is given by $$ \sqrt{x}\stackrel{L_2}{=}\sum_{n\geq 0}\frac{2(-1)^n}{(1-2n)(2n+3)}\,Q_n(x)\tag{3} $$ and obviously $\sqrt{x}$ and $a_0+a_1 x+\ldots+a_n x^n$ are orthogonal on $(0,1)$ iff $$ \sum_{k=0}^{n}\frac{a_k}{k+\frac{3}{2}} = 0.\tag{4}$$

Answer 2

What about $\;f(x)=1-\frac43\sqrt x\;$ ?

Answer 3

You could also try an ansatz of $f(x)=Ax+B$. One such that works is $B=3/2, \; A=-5/2$.

Answer 4

Let $f$ be any continuous function on $[0,1]$ for which $\int_{0}^{1}f(t)\sqrt{t}dt \ne 0$, and which is not a scalar multiple of $\sqrt{x}$. For example, $f(x)=1$. Define a continuous function $g$ on $[0,1]$ by $$ g(x) = f(x)-\frac{\int_{0}^{1}f(t)\sqrt{t}dt}{\int_{0}^{1}tdt}\sqrt{x} $$ Then $g\ne 0$ and $$ \int_{0}^{1}g(x)\sqrt{x}dx =0. $$