1. ## Integrals problem

Let f be a continuous function on open interval I.

Let h>0 and a<b so that: (a-h,b+h) c I.

Now we choose d: 0 < d< h

and now for every x in(a,b) we define the following function:

G(x) = 1/2d * {int(-d --> d) f(x+t)dt}

Prove:
1. G is differentiable functions whose derivative is continuous.
2. lim(d-->0+) G(x) = f(x)

2. Originally Posted by Also sprach Zarathustra
Let f be a continuous function on open interval I.

Let h>0 and a<b so that: (a-h,b+h) c I.

Now we choose d: 0 < d< h

and now for every x in(a,b) we define the following function:

G(x) = 1/2d * {int(-d --> d) f(x+t)dt}

Prove:
1. G is differentiable functions whose derivative is continuous.
2. lim(d-->0+) G(x) = f(x)
Notice that $\frac1\delta\int_d^{d+\delta}\kern-12pt f(x+t)\,dt -f(x+d) = \frac1\delta\int_d^{d+\delta}\kern-12pt \bigl(f(x+t)-f(x+d)\bigr)\,dt \to0$ as $\delta\searrow0$. This is because of the continuity of f, which means that we can take $|f(x+t)-f(x+d)|<\varepsilon$ for all t in the interval $[d,d+\delta]$. So $\int_d^{d+\delta}\kern-12pt \bigl|f(x+t)-f(x+d)\bigr|\,dt < \delta\varepsilon$.

For 1, $\frac{G(x+\delta)-G(x)}\delta = \frac1{2d\delta}\biggl(\int_d^{d+\delta}\kern-12pt f(x+t)\,dt - \int_{-d-\delta}^{-d}\kern-12pt f(x+t)\,dt\biggr) \to \frac1{2d}\bigl(f(x+d)-f(x-d)\bigr)$ as $\delta\searrow0$.
For 2, $G(x) - f(x) = \frac1{2d}\int_{-d}^d\!\!\!\bigl(f(x+t)-f(x)\bigr)\,dt$, which you can estimate in the same way.