Showing Minkowski integral inequality with $p = 2$.

Question

I have shown:

$$\bigg(\int_{0}^{1}f(t)g(t)dt\bigg)^{2} \leq \int_{0}^{1}g(t)^{2}dt\int_{0}^{1}f(t)^{2}dt$$

and now I'd like to use this to show the Minkowski inequality for $p=2$, i.e. $$\Bigg(\int_{0}^{1}(f(t) + g(t))^{2}dt\Bigg)^{\frac{1}{2}} \leq \Bigg(\int_{0}^{1}f(t)^{2} dt\Bigg)^{\frac{1}{2}} + \Bigg(\int_{0}^{1} g(t)^{2}dt\Bigg)^{\frac{1}{2}}.$$

I'm stuck here and I think I'm overthinking things...

$$ \int_{0}^{1}(f(t) + g(t))^{2}dt = \int_{0}^{1}f(t)^{2}dt + 2\int_{0}^{1}f(t)g(t)dt + \int_{0}^{1}g(t)^{2}dt $$

Assuming no knowledge of Hölder's inequality.

Answer 1

By the Cauchy-Scwarz inequality,

$$\left|\int_0^1 f(t)g(t)\, dt\right| = \sqrt{\left(\int_0^1 f(t)g(t)\, dt\right)^2}\le \sqrt{\int_0^1 f(t)^2\, dt}\cdot \sqrt{\int_0^1 g(t)^2\, dt}.$$

Thus

\begin{align}&\int_0^1 (f(t) + g(t))^2\, dt\\ &=\int_0^1 f(t)^2\, dt + 2\int_0^1 f(t)g(t)\, dt + \int_0^1 g(t)^2\,dt \\ &\le \int_0^1 f(t)^2 + 2\sqrt{\int_0^1 f(t)^2\, dt}\cdot \sqrt{\int_0^1 g(t)^2\, dt} + \int_0^1 g(t)^2\, dt\\ &= \left(\sqrt{\int_0^1 f(t)^2\, dt} + \sqrt{\int_0^1 g(t)^2\, dt}\right)^2. \end{align}

Taking square roots, we obtain the result.

Answer 2

From your last line, we have

$$ \int_0^1(f(t) + g(t))^2\,dt = \int_0^1f(t)^2\,dt + 2\int_0^1f(t)g(t)\,dt + \int_0^1g(t)^2\,dt. $$

Now, using the first inequality (the one you say you have shown, which, by the way, is Hölder's inequality) on the middle term on the right hand side, we get

\begin{align} \int_0^1(f(t) + g(t))^2\,dt & \leq \int_0^1f(t)^2\,dt + 2\left(\int_0^1g(t)^2\,dt\int_0^1f(t)^2\,dt\right)^{\frac12} + \int_0^1g(t)^2\,dt = \\ & = \left(\left(\int_0^1f(t)^2\,dt\right)^{\frac12}\right)^2 + 2\left(\int_0^1g(t)^2\,dt\int_0^1f(t)^2\,dt\right)^{\frac12} + \left(\left(\int_0^1g(t)^2\,dt\right)^{\frac12}\right)^2 = \\ & = \left(\left(\int_0^1f(t)^2\,dt\right)^{\frac12} + \left(\int_0^1g(t)^2\,dt\right)^{\frac12}\right)^2. \end{align}

By taking square roots on both sides we get Minkowski inequality.