Solve Inhomogeneous Equation: Show \| u \|_{\infty} \leq \frac{1}{4} \| f \|_1

[tom_rylex]

Homework Statement

I have the solution to an inhomogeneous equation:
$$u(x) = \int_{0}^{1} g(x,t)f(t)dt$$
$$g(x) = x(1-t) , 0<x<t$$
and
$$g(x) = t(1-x), x<t<1$$

Show that
$$\| u \|_{\infty} \leq \frac{1}{4} \| f \|_1$$

Homework Equations

I already know that
$$\| u \|_{\infty} \leq \frac{1}{8} \| f \|_{\infty}$$
and
$$\| u \|_{1} \leq \frac{1}{8} \| f \|_{1}$$

The Attempt at a Solution

I think I'd like to say that
$$sup| f | \leq 2| f |$$
for some x in f, and
$$\| u \|_{\infty} \leq 2*\| u \|_1 \leq 2*\frac{1}{8} \| f \|_1$$
Therefore
$$\| u \|_{\infty} \leq \frac{1}{4} \| f \|_1$$
Is this adequate, or do I need to say something more to complete the proof?

 

[mighty2000]

Your solution is on the right track, but there are a few things that could be clarified to make it more complete. First, you should explain why the inequality \| u \|_{\infty} \leq \frac{1}{8} \| f \|_{\infty} is true in the first place. This can be done by using the properties of the integral and the given functions g(x,t) to show that the maximum value of u(x) occurs when f(t) is at its maximum value, which is \| f \|_{\infty}.

Next, you should explain how you obtained the inequality \| u \|_{1} \leq \frac{1}{8} \| f \|_{1}. This can be done by using the triangle inequality and the given functions g(x,t) to show that the integral of u(x) is less than or equal to the integral of f(t) over the same interval.

Finally, you can use the inequalities \| u \|_{\infty} \leq \frac{1}{8} \| f \|_{\infty} and \| u \|_{1} \leq \frac{1}{8} \| f \|_{1} to get the desired result \| u \|_{\infty} \leq \frac{1}{4} \| f \|_1 by using the fact that \| f \|_{\infty} \leq \| f \|_{1}.

Overall, your solution is on the right track but could use some additional explanation and justification to make it more complete.