how to prove $\|T-T_n\|\le n^{-1/2}$, and is $T$ compact?

Question

suppose $x\in [0,1]$,$$Tf(x)=\int_0^xf(t) \, dt,\\T_nf(x)=\sum_0^{n-1} X_{[{k\over n}, {{k+1}\over n})}(x) \int_0^{k\over n} f(y) \, dy$$,where we have $X_{[{k\over n},{{k+1}\over n})}(x)=1$ if $x\in{[{k\over n},{{k+1}\over n})},$ otherwise $0$.

(1) $\|T-T_n\|\le n^{-1/2}$

(2) prove that $T$ is compact (here is the definition of compact operator:https://en.wikipedia.org/wiki/Compact_operator)

(notice that $T$ is a linear and continuous map from $L^2[0,1]\to L^2[0,1]$)

I don't how to prove the questions above, can someone help me prove it?

For $x \in [k/n, (k+1)/n)$, can you explicitly determine $Tf(x) - T_nf(x)$? — Daniel Fischer, Nov 25 '16 at 16:52
It is smaller than or equal to $|\int_{k/n}^{{k+1}/n}f(x)dx|$(1), but the thing I am confused about is that I need to prove $||(T-T_n)(f)||<=n^{-1/2}||f||$ (2)for any f ,but I am not sure how equation (1) is related to (2) — user144600, Nov 25 '16 at 17:05

Tsemo Aristide · Answer 1 · 2016-11-25T17:51:15.097

0

Hint: To show $(1)$, use the fact that $\int_{k/n}^{(k+1)/n}f(y)dy=f(y_k)/n, y_k\in [k/n,(k+1)/n]$, then find a bound of $| T(f)(x)-T_n(f)(x)|$.

The image of $T_n$ is in a finite dimensional vector space, (1) implies that $T$ is the limit of $T_n$ so is compact by definition.

edited Nov 25 '16 at 17:51

answered Nov 25 '16 at 15:19

Tsemo Aristide

87,475

How can you get $\int_{k/n}^{(k+1)/n}f(y)dy=f(y_k)/n, y_k\in [k/n,(k+1)/n]$? – user144600 Nov 25 '16 at 15:30
I assume that $f$ is continue, – Tsemo Aristide Nov 25 '16 at 15:32
but I think f is just in $L^2[0,1]$, I thinkI understand why T is compact now, but as for question(1), still confused – user144600 Nov 25 '16 at 15:48
Then use the fact that continue functions are dense in $L^2[0,1]$. http://math.stackexchange.com/questions/226049/continuous-functions-on-0-1-is-dense-in-lp0-1-for-1-leq-p-infty – Tsemo Aristide Nov 25 '16 at 17:06

score 0 · Accepted Answer · answered Nov 25 '16 at 21:29

Let's first look at a fixed $n$. For $x \in \bigl[\frac{k}{n}, \frac{k+1}{n}\bigr)$ we have

$$T_n f(x) = \int_0^{k/n} f(t)\,dt,$$

and therefore

$$Tf(x) - T_nf(x) = \int_{k/n}^x f(t)\,dt.\tag{1}$$

To get a nice bound for $\lvert Tf(x) - T_nf(x)\rvert$ we use the Cauchy-Schwarz inequality,

$$\lvert Tf(x) - T_nf(x)\rvert \leqslant \sqrt{\int_{k/n}^x 1^2\,dt} \cdot \sqrt{\int_{k/n}^x \lvert f(t)\rvert^2\,dt} \leqslant \sqrt{x - \tfrac{k}{n}} \cdot \lVert f\rVert_{L^2([0,1])}.\tag{2}$$

Since $x - \frac{k}{n} < \frac{k+1}{n} - \frac{k}{n} = \frac{1}{n}$, we see that $\lvert Tf(x) - T_nf(x)\rvert \leqslant \frac{1}{\sqrt{n}}\lVert f\rVert_{L^2([0,1])}$ for all $x \in [0,1]$ (the equality $T_nf(1) = Tf(1)$ follows directly from the definitions). Thus we find

$$\lVert (T - T_n)f\rVert_{L^2([0,1])}^2 = \int_0^1 \lvert Tf(x) - T_nf)(x)\rvert^2\,dx \leqslant \int_0^1 \frac{1}{n}\lVert f\rVert_{L^2([0,1])}^2\,dx = \frac{1}{n}\lVert f\rVert_{L^2([0,1])}^2,$$

from which we obtain $\lVert T - T_n\rVert \leqslant n^{-1/2}$ by taking the square root on both sides, and then the supremum over all $f$ with $\lVert f\rVert_{L^2([0,1])} \leqslant 1$. (Aside: if we had used the right hand side of $(2)$ rather than the bound $\sqrt{x - \frac{k}{n}} \leqslant \frac{1}{\sqrt{n}}$ we would have obtained the slightly sharper bound $\lVert T - T_n\rVert \leqslant \frac{1}{\sqrt{2n}}$.)

Now $T_n$ is a continuous linear operator with finite-dimensional range - the range of $T_n$ is spanned by the characteristic functions of the intervals $\bigl[\frac{k}{n}, \frac{k+1}{n}\bigr)$, $0 \leqslant k < n$ - hence compact. And by $\lVert T - T_n\rVert \to 0$, $T$ is the norm-limit of a sequence of compact operators. Since the subspace of compact operators is closed, it follows that $T$ is compact too.

It is brilliant! Thanks for help! – user144600 Nov 25 '16 at 23:01 — user144600, Nov 25 '16 at 23:01

how to prove $\|T-T_n\|\le n^{-1/2}$, and is $T$ compact?

2 Answers2