Proving T is a contraction

Question

Trying to sure up some holes in my proof for the following:

Let $I=[0,1]$ and define $T:C(I)\rightarrow C(I)$ by $$(Tf)(x)=x+\int_{0}^{x}(x-t)f(t)dt \quad \bigl(x\in I, f\in C(I)\bigr).$$

Show that $T$ is a contraction on $C(I)$ (with uniform metric).

$$\begin{align} \bigl\lvert(Tf)(x)-(Tg)(x)\bigr\rvert &= \biggl\lvert x+\int_{0}^{x}(x-t)f(t)dt-x+\int_{0}^{x}(x-t)g(t)dt\biggr\rvert \\ &= \biggl\lvert\int_{0}^{x}(x-t)f(t)dt-\int_{0}^{x}(x-t)g(t)dt\biggr\rvert \\ &\le d_\infty(f,g)\int_{0}^{x}(x-t)dt$ \end{align}$$ [can someone explain succinctly why this particular step is true].

Clearly, $\int_{0}^{x}(x-t)dt$ evaluates to a number between 0 and 1/2 (since $x\in I$) leading to $\lvert(Tf)(x)-(Tg)(x)\rvert\le \frac{1}{2}d_\infty(f,g)\Longrightarrow T$ is a contraction on $C(I)$.

Does this look to be adequate?

By definition, we know that T has a unique fixed point. My book claims it is $\sinh(x)$. How do I show this?

Answer 1

The step in question is why does it follow that \begin{align*} \left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) g(t) dt \right | \leq d_\infty(f,g) \int_0^x (x-t) dt. \end{align*} This follows easily from the defintion of $d_\infty(f,g) = \sup_{s \in [0,1]}|f(s) - g(s)|$ and the the estimate \begin{align*} \left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) g(t) dt \right | &= \left | \int_0^x (x-t) [f(t) - g(t)] dt\right | \\ &\leq \int_0^x (x-t) |f(t) - g(t)| dt \\ &\leq \int_0^x (x-t) \sup_{s \in [0,1]}|f(s) - g(s)| dt \\ &= d_\infty(f,g) \int_0^x (x-t)dt. \end{align*} To see that the fixed point $f(x) = \sinh x$, observe that a fixed point satisfies (by definition) \begin{align*} f(x) = x + \int_0^x (x-t) f(x) dx. \end{align*} Differentiating the above expression twice and using the fundamental theorem of calculus yield \begin{align*} f''(x) = f(x). \end{align*} Moreover, $f(0) = 0$, and $f'(0) = 1$. The general solution to the above ODE is $f(x) = a e^x + b e^{-x}$ for some constants $a,b$. The initial conditions $f(0) = 0$, $f'(0) = 1$ imply that $a = 1/2$, $b = -1/2$ so that \begin{align*} f(x) = \frac{1}{2} e^x - \frac{1}{2} e^{-x} = \sinh x. \end{align*}

Answer 2

Succinctly, we have for $x\in I=[0,1]$

$$\begin{align} \left|\int_0^x (x-t)\left(f(t)-g(t)\right)\,dt\right|&\le\int_0^x (x-t)\left|f(t)-g(t)\right|\,dt \tag 1\\\\ &\le \sup_{t\in [0,x]}\left|f(t)-g(t)\right|\int_0^x (x-t)\,dt \tag 2\\\\ &\le\frac12\,\sup_{t\in [0,x]}\left|f(t)-g(t)\right| \tag 3 \end{align}$$

since $f$ and $g$ (and hence $f-g$) are in $C(I)$.

In arriving at $(1)$, we used the Triangle inequality.

In going from $(1)$ to $(2)$, we used the MVT for integrals.

In arriving at $(3)$, we used the fact the $\int_0^x(x-t)\,dt=\frac{x^2}{2}\le\frac12$.

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To show that $\sinh x$ is a fixed point, we see that

$$\begin{align} x+\int_0^x (x-t)\sinh t\,dt&=x+x\left.\cosh t\right|_0^x-\left.(t\cosh t-\sinh t)\right|_0^x\\\\ &=\sinh x \end{align}$$

and we are done!

Answer 3

We have the inequality $$ \left\lvert \int_a^b f \right\rvert \leqslant \int_a^b \lvert f \rvert, $$ basically as a generalisation of the triangle inequality. Applying this to your integral gives $$ \left\lvert \int_0^x (x-t) (f(t)-g(t)) \, dt \right\rvert \leqslant \int_0^x (x-t) \lvert f(t)-g(t) \rvert \, dt. $$

If $u$ is a nonnegative function on $(a,b)$ and $\lvert v(x) \rvert \leqslant M$ for almost every $x$ in $(a,b)$, then $ -Mv(x) \leqslant u(x)v(x) \leqslant Mv(x) $ for almost every $x$ in $(a,b)$ and hence, integrating this inequalities, $$ -M\int_a^b u(x) \, dx \leqslant \int_a^b u(x)v(x) \, dx \leqslant M\int_a^b u(x) \, dx. $$

You can use this on your integral, with $u(t)=x-t$ and $v(t)=\lvert f(t)-g(t) \rvert$, to get to your $d_{\infty}(f,g) \int_0^x (x-t) \, dt$ (here $M$ being the distance $d_{\infty}(f,g)$, of course).

The rest looks fine. As far as finding the fixed point goes, perhaps you'll spot it more easily in differential form: note that if $$ f(x) = x + \int_0^x (x-t)f(t) \, dt, $$ then $f(0)=0$, $f'(0) = 1+0=1$, and differentiating, $$ f'(x) = 1 + (x-x)f(x) + \int_0^x f(t) \, dt = 1 + \int_0^x f(t) \, dt \\ f''(x) = f(x) $$ Now, by the uniqueness theorem for ODEs, the only solution to these equations is $\sinh{x}$, as required.

Answer 4

You have $$\left|\int_0^x (x - t) (f(t) - g(t))\ dt \right|.$$ I assume your notation $d_\infty(f, g)$ stands for the $L^\infty$ norm $\|f - g\|_\infty$. In this case I can shed some light on the next step. Assume for convenience that $f$ and $g$ are continuous (the next step holds regardless, I just think it's more intuitive if they are). Then we can crash through with the absolute values to get $$\left|\int_0^x (x - t) (f(t) - g(t))\ dt \right| \le \int_0^x \left| (x - t) \right| \left| f(t) - g(t) \right|\ dt\ .$$ But the second integrand we can bound by $|f(t) - g(t)| \le \max_{t\in (0,x)} |f(t) - g(t)| = \| f - g \|_\infty$. Thus $$ \int_0^x \left| (x - t) \right| \left| f(t) - g(t) \right|\ dt \le \|f - g\|_\infty \int_0^x (x - t)\ dt\ .$$ I dropped the absolute values on the integrand because it is clearly non-negative.

Everything else looks good.

To show that $\sinh(x)$ is a fixed point, simply set $f(x) := \sinh(x)$ and evaluate $\left(Tf\right)(x)$. Presumably, you'll get $\sinh(x)$ as your answer.