Showing that $L^2\subset L^1$ for $L^2([0,t_f])$, with $t_f$ a fixed positive number.

Question

I saw demonstrations using the Cauchy-Schwarz Inequality but I am still not convinced because the Inequality is as follows :

$$ \left |\langle f,g\rangle\right | \leq \left \|f \right \|_{L_2} . \left \|g \right \|_{L_2} $$

And if we consider f to be L2 and if we take g = 1 we have : $$ \left | \int_{0}^{t_f}f(t)dt\right | \leq \left \|f \right \|_{L_2} . \left \|1 \right \|_{L_2} $$

I feel confused because we have : $$ \left | \int_{0}^{t_f}f(t)dt\right | \leq \int_{0}^{t_f} |f(t)|dt $$

And thus we have no comparison relationship between $$\int_{0}^{t_f} |f(t)|dt$$ and $$\left \|f \right \|_{L_2}$$ which is supposed to be finite.

Answer 1

Try applying Cauchy-Schwarz to $|f|$ and $1$.

Answer 2

In terms of getting a proof done, it really is just Cauchy-Schwarz:

$$\left | \int_0^t f(s) ds \right | = (f,1)_{L^2} \leq \| f \|_{L^2} \| 1 \|_{L^2} = t^{1/2} \left ( \int_0^t f(s)^2 ds \right )^{1/2}.$$

Cauchy-Schwarz can be proven in the abstract setting of a general inner product space, and then it follows in your case from the fact that the $L^2$ inner product is an inner product.

In terms of intuition, you may want to look into Jensen's inequality. The most basic form says that if $f$ is convex and $a_i \geq 0$ are such that $\sum_{i=1}^n a_i = 1$, then

$$f \left ( \sum_{i=1}^n a_i b_i \right ) \leq \sum_{i=1}^n a_i f(b_i).$$

Since $f(x)=x^2$ is convex, you can work a little to get the corresponding result for integrals from $0$ to $1$ (which are in a sense convex combinations, as you can see by writing Riemann sums), and then a scaling argument gets you the factor of $t^{1/2}$ that we had above.