I study my linear algebra midterm. I saw a question in the problem set but I could not solve the question. I know that I should show the followings but I don't know how I can start to the solution.
i) $\|v\| \ge 0$ and $\|v\| = 0$ if $v = 0$,
ii) $\|av\| = |a|\cdot \|v\|$,
iii) $\|v_1 + v_2\| \le \|v_1\| + \|v_2\|$ (Triangle inequality).
Here is the question, I need hints and techniques for the solution. (Not necessarily for all of them, I can handle if I learn what should I do). Thanks.
Consider the set of functions $F = \{ f : \Bbb{R} \to \Bbb{R} : f(t) = a \cos(t) + b \sin(t), a, b \in \Bbb{R}\}$ which is a linear space over the field of real numbers. For each of the below cases state whether the candidate is a norm or not.
(a) $\|f\| = a^2 + b^2$
(b) $\|f\| = |a| + |b|$
(c) $\|f\| = |a + b|$
(d) $\|f\| = |a + b| + |b|$
(e) $\|f\| = \max_{t \in \Bbb{R}}|f(t)|$.
You need to verify the three conditions you mentioned. For each case, in general, the first conditions are easy to verify, while the third one may be a little tricky. I will help you with the last case.
First you need to verify that $\forall f\in F,\lVert f \rVert \geq 0$ and that it is equal to zero exactly when $f=0$. Now, $\lVert f \rVert = \max \lvert f(t) \rvert$ is clearly non-negative since $\forall t,\lvert f(t) \rvert \geq 0$. If $\max \lvert f(t) \rvert =0$, then $\forall t, f(t)=0$, in particular for $t=0,t=\frac{\pi}{2}$..., so $a=b=0$, hence $f=0$.
Let $c\in \mathbb{R}$, $$\lVert cf \rVert=\max \lvert c f(t) \rvert=\max \lvert c \rvert\lvert f(t) \rvert=\lvert c \rvert \max \lvert f(t) \rvert=\lvert c \rvert \lVert f \rVert,$$ so the second condition holds.
Let $f,g \in F$, then \begin{align*} \lVert f+g \rVert&=\max \lvert f(t)+g(t) \rvert \\ &\leq \max (\lvert f(t) \rvert+\lvert g(t) \rvert) \\ &\leq \max \lvert f(t) \rvert+\max\lvert g(t) \rvert) \\ &\leq \lVert f \rVert + \lVert g \rVert. \end{align*} So the fourth option indeed defines a norm on $F$.