limit of integral of series of functions

Question

helo,

I don't know how to solve this one:

$f_n(x)$ is a series of continuous functions in [a,b] that is uniformly converge to f(x) in [a,b]. prove that for every p>0 $\lim_{n \to \infty} \int_a^b|f(x)-f_n(x)|^pdx=0$

i'll explain what i tried:

since $f_n(x)$ is continuous on [a,b] and is uniformly convergent to f(x), i think that the following property can be derived:$lim_{n \to \infty} \int_a^bf_n(x)dx=\int_a^b f(x)dx$. however i don't know how to continue and apply it because of the absolute value and $^p$(not that $^p$ should matter because it's a scalar and as long it's positive it won't interfer with the uniform convergence of $f_n(x)$ to f(x)$ in this domain([a,b].

don't know how to continue or prove it.

thank you very much for helping.

Answer 1

If $(f_n)$ converges uniformly to $f$ on $[a,b]$, then the sequence $(g_n)$, where $g_n:=|f_n-f|^p$, converges uniformly to $0$, hence

$$\lim_{n \to \infty} \int_a^bg_n(x) dx=0.$$

Answer 2

Other way to see this : )

Notation: X is the interval [a,b], $\mu(X)=b-a$, $f_n \rightarrow_{L^p} f$ the desired convergence and $f_n \rightarrow_u f$ the uniform convergence.

Proof:

Let $ \varepsilon>0$, we can take $\delta>0$ such that $\delta . [\mu(X) ]^{\frac{1}{p}} < \varepsilon$, $$\delta < \frac{\varepsilon}{ [\mu(X) ]^{\frac{1}{p}} }. $$ Given $f_n \rightarrow_u f$, exists $n_0 \in N$, such that for every $n \geq n_0$ we have

$$ |f_n(x)-f(x)|\leq \delta\; \forall x \in X . $$ Then, we have $$||f_n-f||_{L^p}=\left(\int_X |f_n-f|^pd\mu \right)^{\frac{1}{p}} \leq \left(\int_X \delta^p d\mu \right)^{\frac{1}{p}} =\delta [\mu(X) ]^{\frac{1}{p}} < \varepsilon . $$

So $f_n \rightarrow_{L^p} f.$