Show that C'[a, b] is subspace of C [a, b]

Show that C'[a, b] is subspace of C [a, b]
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Cristiano Sears
To shiw that C'[a.b] is subspace of C[a,b]. Suppose ${f}_{1}$ and ${f}_{2}$ are in C'[a.b]. We will prove that ${f}_{1}+{f}_{2}$ is in C'[a,b]. Let $f={f}_{1}+{f}_{2}$. Then from the definition of derivative we have
$\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\left(\frac{{f}_{1}\left(x+h\right)-{f}_{1}\left(x\right)}{h}\right)+\frac{{f}_{2}\left(x+h\right)-{f}_{2}\left(x\right)}{h}\right)\right)=\left(\underset{h\to 0}{lim}\left(\frac{{f}_{1}\left(x+h\right)-{f}_{1}\left(x\right)}{h}\right)\right)+\underset{h\to 0}{lim}\left(\frac{{f}_{2}\left(x+h\right)-{f}_{2}\left(x\right)}{h}\right)\right).$
Whis last equality is true if both of the limits on the last line exist, and it is given that they do. Saying the lat equality is true means the limits involved exist. The relevant theorem is that if two limits as $h\to$ something equal to the sum of those two limits.
The first equality above is just algebra: adding factions. Nowto show for $c\in \mathbb{R}$, $c{f}_{1}\in {C}^{\prime }\left[a,b\right]$. We know that
$\underset{h\to 0}{lim}\frac{c{f}_{1}\left(x+h\right)-cf\left(x\right)}{h}=c\underset{h\to 0}{lim}\frac{{f}_{1}\left(x+h\right)-{f}_{1}\left(x\right)}{h}$
By similar argument, since limit of right side part of above equation exists so, $c{f}_{1}$ is also differentiable, therefore, $c{f}_{1}\in {C}^{\prime }\left[a,b\right]$. Hence C'[a,b] is a subspace of C[a,b].