# Determine all real possible s, so that W={f|int_-1^1f(x)dx=s} is subspace in real function that continue in [-1,1] ?

## $W = \left\{f | {\int}_{-} {1}^{1} f \left(x\right) \mathrm{dx} = s\right\}$

Apr 10, 2017

$0$

#### Explanation:

If ${f}_{1} \in {W}_{s}$ then ${\int}_{a}^{b} {f}_{1} \mathrm{dx} = s$ and

If ${f}_{2} \in {W}_{s}$ then ${\int}_{a}^{b} {f}_{2} \mathrm{dx} = s$

but if ${W}_{s}$ has structure of subspace then

$\alpha {f}_{1} + \beta {f}_{2} = {\int}_{a}^{b} \left(\alpha {f}_{1} + \beta {f}_{2}\right) \mathrm{dx} = \alpha {\int}_{a}^{b} {f}_{1} \mathrm{dx} + \beta {\int}_{a}^{b} {f}_{2} \mathrm{dx} = \left(\alpha + \beta\right) s$

so the only feasible value for $s$ is $0$

Now, with $s = 0$, if ${f}_{1} \in {W}_{0}$ and ${f}_{2} \in {W}_{0}$ then

$\alpha {f}_{1} + \beta {f}_{2} \in {W}_{0}$

NOTE:

With $a = b = 1$ all odd functions (such that $f \left(x\right) = - f \left(- x\right)$) are contained in ${W}_{0}$