Let $f_n \in C([0,2014]) $ Show, that if $f_n \rightrightarrows f$ and for all n $\int_{[0,2014]} ff_n dl_1=0$ then $ f\equiv 0$ Let $f_n \in C([0,2014]) $  Show, that if $f_n \rightrightarrows f$ and for all n $\int_{[0,2014]} ff_n dl_1=0$ then $ f\equiv 0$
I have no idea how to start the exercise like that.

Assume that $f$ and all $f_n$ are real-valued.
We can show that $f_n\rightrightarrows f$ implies $$\int_0^{2014}f_n^2(x)dx\to\int_0^{2014}f ^2(x)dx. $$
Indeed, $$\left|\int_0^{2014}(f_n^2(x)-f ^2(x))dx\right|\le 2014\cdot \|f_n+f\|_\infty\cdot\|f_n-f\|_\infty. $$ The second factor is bounded (and converges to $2\|f\|_\infty$), the third one converges to zero.
Now let's look at  $$\int_0^{2014}(f_n(x)+f(x))^2dx.$$
On the one hand , it is equal to $$\int_0^{2014}(f_n^2(x)+f^2(x))dx $$ by assumption and converges therefore to $$2\int_0^{2014} f^2(x)dx.$$
On the other hand, $$\left|\int_0^{2014}\left((f_n(x)+f(x))^2-4f^2(x)\right)dx\right|\le 2014\cdot \|f_n+3f\|_\infty\cdot\|f_n-f\|_\infty \to 0.$$
So, the integral $\int_0^{2014}(f_n^2(x)+f^2(x))dx$ converges to both $2\int_0^{2014} f^2(x)dx $ and $4\int_0^{2014} f^2(x)dx $, which results in only one option: 
$$2\int_0^{2014} f^2(x)dx =0.$$
This an integral of a non-negative function, so we can conclude that this function is zero almost everywhere. It's a continuous function, hence it's zero everywhere.
