Let \((X,\mcm,\mu)\) be a measure space, let \(f, g\in L^2(\mu)\), and put \(\inner{f}{g}\coloneqq\int_X f\overline{g}\dmu\). Then \(fg\in L^1(\mu)\) and \[ \abs{\inner{f}{g}} \leq \norm{fg}_1 \leq \norm{f}_2 \norm{g}_2, \quad\text{that is,}\quad \abs{\int_X f\overline{g}\dmu}\leq\int_X\abs{fg}\dmu \leq \qty{\int_X \abs{f}^2\dmu}^{1/2} \qty{\int_X \abs{g}^2\dmu}^{1/2}. \] Moreover, \(\abs{\inner{f}{g}} = \norm{f}_2\norm{g}_2\) if and only if \(f\) and \(g\) are linearly dependent in \(L^2(\mu)\).
The first inequality is \(\abs{\int_X f\overline g\dmu}\leq\int_X\abs{f\overline g}\dmu\). For the second, if \(\norm{f}_2 = 0\) or \(\norm{g}_2 = 0\) then \(fg = 0\) almost everywhere and both sides vanish. Otherwise, applying \(ab\leq\frac12(a^2+b^2)\) for \(a,b\geq 0\) to \(a = \abs{f(x)}/\norm{f}_2\) and \(b = \abs{g(x)}/\norm{g}_2\) and integrating gives \(\int_X\abs{fg}\dmu\leq\norm{f}_2\norm{g}_2\).
For the equality case, if \(f = \lambda g\) or \(g = \lambda f\) for some \(\lambda\in\CC\), then both sides equal \(\abs{\lambda}\norm{g}_2^2\) or \(\abs{\lambda}\norm{f}_2^2\). Conversely, suppose \(\abs{\inner{f}{g}} = \norm{f}_2\norm{g}_2\). If \(\norm{g}_2 = 0\) then \(g = 0\) in \(L^2(\mu)\) and \(f, g\) are dependent. Otherwise put \(\lambda\coloneqq\inner{f}{g}/\norm{g}_2^2\); expanding gives \[ \norm{f-\lambda g}_2^2 = \norm{f}_2^2 - \frac{\abs{\inner{f}{g}}^2}{\norm{g}_2^2} = 0, \] so \(f = \lambda g\) in \(L^2(\mu)\).