Sup norm can always be made smaller without altering the value of the Riemann-integral

Set $\alpha>0$ and let $V := \{f\in C^0([0,1]) : f(0) = 0\}$ be equipped with the sup norm $\Vert\cdot \Vert_\infty$. I'm trying to show that for every $f\in V$ with $\int_{0}^{1} f(x) \, dx = \alpha$ there is $g\in V$ such that $\int_{0}^{1} g(x) \, dx = \alpha$ and $\Vert g\Vert_\infty < \Vert f\Vert_\infty$. Intuitively I feel that this should be achievable because one must "make $f$ smaller" at the points where it reaches its maximum and "make $f$ larger" at certain places where it does not reach its maximum to compensate for the value of the integral. However, I don't really have a clear and formal idea of ​​how one can do this.