Triangle inequality for an integral

Question

I have the following here.

Fix real numbers $a<b$, let $M$ be a positive real number, and define $$B(M)=\left\{ f \in C[a,b] \bigg| \int_{a}^{b}f(x)^2 dx \leq M \right\}$$ Show that if $f,g\in B(M)$, then $tf+(1-t)g \in B(M)$ for all $t \in[0,1]$.

I'm given a hint on using the triangle inequality for the functions $tf$ and $(1-t)g$. I'm not really sure how to proceed though.

The triangle inequality says: $||\vec{u}+\vec{v}||\leq||\vec{u}||+||\vec{v}||$

I am not sure what to do here though. Can someone guide me through how I would proceed?

Answer 1

Let $h:=tf+(1-t)g$. Then $h^2=t^2f^2+2t(1-t)fg+(1-t)^2g^2.$ Thus

$$\int_a^b h^2 dx \le t^2M+2t(1-t)\int_a^b fg dx +(1-t)^2 M.$$

By Cauchy-Schwarz

$$\int_a^b fg dx \le (\int_a^b f^2 dx)^{1/2} \cdot (\int_a^b g^2 dx)^{1/2} \le \sqrt{M} \sqrt{M}=M.$$

Therefore

$$\int_a^b h^2 dx \le t^2M+2t(1-t)M+(1-t)^2M=M.$$

Consequence: $h \in B(M).$

Remark: similar computations show that if $ts \ge 0$ and $|t+s| \le 1,$ then $tf+sg \in B(M).$