Show that $\|f\|=\left(\int_0^1 (|f|^2+|f'|^2)dx\right)^{(1/2)}$ is a norm in $C^1([0,1])$ Let $C^1([0,1])$ be the space of all functions having continuous derivative. For each $f\in C^1([0,1])$, set
$$\|f\|=\left(\int_0^1 (|f|^2+|f'|^2)dx\right)^{(1/2)}$$
Show that $\|\cdot\|$ is a norm of the space $C^1([0,1])$.


How do I solve this?

Is this as simple as showing I have absolute homogeneity, triangular inequality and a zero vector? As well as representing my space as a vector space?


Will I have to prove all of the axioms one by one of a vector space, to show this is a vector space?


For showing it is a norm, satisfying the absolute homogeneity is quite easy:
$$\|cf\|=\left(\int_0^1 (|cf|^2 + |cf'|^2) dx\right)^\frac12 = \left(\int_0^1 (c^2|f|^2+c^2|f'|^2)dx\right)^\frac12$$
$$=\left(c^2\int_0^1 (|f|^2+|f'|^2)dx\right)^\frac12=|c|\|f\|$$
 A: The set of real functions on the interval $[0,1]$ is certainly a vector space under the operations
$$
f+g\colon x\mapsto f(x)+g(x),\qquad
\alpha f\colon x\mapsto \alpha f(x)
$$
for $f$ and $g$ any real functions on $[0,1]$ and $\alpha$ any real number. Proving the axioms is very easy.
The set $C^1([0,1])$ is easily seen to be a subspace of the above vector space, because derivation is linear and linear combinations of continuous functions are continuous.
Now you have to prove the properties of a norm.


*

*“$\|f\|\ge0$” is obvious

*“$\|f\|=0$ implies $f=0$” is easy: if a nonnegative continuous function has zero integral, then it is the constant zero function

*Homogeneity is easy as well and indeed your proof is correct.


As usual, the toughest part is the triangle inequality. This can be obtained from the Cauchy-Schwarz inequality if you are able to connect this (yet to be proved) norm with an inner product.
Define, for any two continuous real function $f$ and $g$ on $[0,1]$,
$$
\langle f,g\rangle_0=\int_0^1 f(x)g(x)\,dx
$$
Then it is easy to show that this defines an inner product on the space $C^0([0,1])$. For functions $f,g\in C^1([0,1])$, define
$$
\langle f,g\rangle_1=\langle f,g\rangle_0+\langle f',g'\rangle_0
$$
and prove that this defines an inner product on $C^1([0,1])$; for instance, if $\langle f,f\rangle_1=0$, then
$$
\langle f,f\rangle_0+\langle f',f'\rangle_0=0
$$
and so $\langle f,f\rangle_0=0$, which already implies $f=0$. The other properties are easy to derive.
Now, the norm associated to this inner product is
\begin{align}
\|f\|&=(\langle f,f\rangle_1)^{1/2}\\
&=(\langle f,f\rangle_0+\langle f',f'\rangle_0)^{1/2}\\
&=\left(\int_0^1 f(x)^2\,dx+\int_0^1 f'(x)^2\,dx\right)^{\!1/2}\\
&=\left(\int_0^1 (f(x)^2 + f'(x)^2)\,dx\right)^{\!1/2}
\end{align}
which is precisely your definition. The general theory about inner products and the induced norm shows what you want. Note that you don't even have to prove points 1., 2. and 3. above, because they are already implicit in the proof that every inner product $\langle \cdot,\cdot\rangle$ on a vector space $V$ defines a norm by setting
$$
\|v\|=(\langle v,v\rangle)^{1/2}
$$
A: You have to prove that for all $f,g\in \mathcal C^1([0,1])$ and all $\lambda\in\mathbb R$ that:
$\|f\|=0\implies f\equiv 0$
$\|\lambda f\|=|\lambda|\|f\|$
$\|f+g\|\leq\|f\|+\|g\|$
