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## Homework Statement

C[a,b] is a vector space of continuos real valued functions. for f,g in C[a,b]

<f,g>=∫f(x)g(x)dx, [a,b]

Give a completely rigorous proof that if <f,f>=0, then f=0

**2. The attempt at a solution**

I tried to prove this by contrapositive, "f≠0 implies that <f,f>≠0

When f(x)≠0, f(x)<0 or f(x)>0. However, [f(x)]

^{2}>0 in both cases. Therefore, there exist a<c<b and a real number d such that [f(c)]

^{2}=d > 0

Since f is a continuous function, f

^{2}is also a continuos function.

**By definition of continuity,**

For all ε>0, exist δ>0 such that

for all x : |x-c|<δ implies that |f

For all ε>0, exist δ>0 such that

for all x : |x-c|<δ implies that |f

^{2}(x)-f^{2}(c)|<εWe choose ε<d/2 and get that

-d/2<f

^{2}(x)-f

^{2}(c)<d/2

0<d/2<f

^{2}(x)<3d/2

So, for all x that is in the interval [c-δ,c+δ], f

^{2}(x)>0

**(can we conclude anything about the x not in this interval? like they'll be 0?)**

**The following is the part where I'm not sure...**

To compute the integral ∫[itex]^{a}_{b}[/itex]f

^{2}(x)

we divide [a,b] into n-subinterval

a=x

_{1}<x

_{2}<...<x

_{i-1}<c-δ<c+δ<x

_{i+1}<...<x

_{n}=b

Let x[itex]^{i}_{*}[/itex]= midpoint of each interval [x[itex]_{i}[/itex],x[itex]_{i+1}[/itex]] where we denote c-δ<x<c+δ

∫[itex]^{a}_{b}[/itex]f

^{2}(x)≈f[itex]^{2}[/itex](x[itex]_{1}[/itex])(x

_{2}-x

_{1})+...+f[itex]^{2}[/itex](x)δ+...+f[itex]^{2}[/itex](x[itex]_{n-1}[/itex])(x

_{n}-x

_{n-1})

>f[itex]^{2}[/itex](x)δ>0

**I'm really not sure about this step**

So since f≠0 implies <f,f>=∫[itex]^{a}_{b}[/itex]f

^{2}(x)≠0.<f,f>=0 implies f=0

Thanks